Research ArticleThe Applications of Cardinal Trigonometric Splines inSolving Nonlinear Integral Equations

Jin Xie1 Xiaoyan Liu2 and Lixiang Xu1

1 Department of Mathematics and Physics Hefei University Hefei 230601 China2Department of Mathematics and Physics University of La Verne La Verne CA 91750 USA

Correspondence should be addressed to Xiaoyan Liu xliulaverneedu

Received 3 December 2013 Accepted 15 January 2014 Published 4 March 2014

Academic Editors Y M Cheng and L You

Copyright copy 2014 Jin Xie et al This is an open access article distributed under the Creative Commons Attribution License whichpermits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

The cardinal trigonometric splines on small compact supports are employed to solve integral equations The unknown function isexpressed as a linear combination of cardinal trigonometric splines functionsThen a simple system of equations on the coefficientsis deducted When solving the Volterra integral equations the system is triangular so it is relatively straight forward to solve thenonlinear system of the coefficients and a good approximation of the original solution is obtained The sufficient condition for theexistence of the solution is discussed and the convergence rate is investigated

1 Introduction

Trigonometric splines were introduced by Schoenberg in [1]Univariate trigonometric splines are piecewise trigonometricpolynomials of the form

119899

sum119896=0

119886119896cos (120572

119896119909) + 119887

119896sin (120572

119896119909) (where 119887

0= 0) (1)

(where 1205720= 0 lt 120572

1lt 1205722lt sdot sdot sdot lt 120572

119899are real numbers)

in each interval and they are nature extensions of polynomialsplines Needless to say trigonometric splines have their ownadvantages A number of papers have appeared to study theproperties of the trigonometric splines and trigonometric B-splines (cf [2ndash4]) since then

In my previous papers (cf [5ndash7]) low degree orthonor-mal spline and cardinal spline functions with small compactsupports were constructed The method can be extendedto construct higher degree orthonormal or cardinal splinesUnlike in the book (cf [1]) by the cardinal splines wemean the specific splines satisfying cardinal interpolationconditions which means that the cardinal function has thevalue one at one interpolation point and value zero at all otherinterpolation points Cardinal splines are not only useful ininterpolation problems but they are also useful in deduction

of numerical integration formulas [6] and in solving integralequations

Integral equations appear in many fields includingdynamic systems mathematical applications in economicscommunication theory optimization and optimal controlsystems biology and population growth continuum andquantum mechanics kinetic theory of gases electricity andmagnetism potential theory and geophysics Many differen-tial equations with boundary value can be reformulated asintegral equations There are also some problems that can beexpressed only in terms of integral equations

In this paper we focus on the Volterra integral equationsof the second kind

119910 (119909) = 119892 (119909) + 120582int119909

119886

119870 (119909 119905) 119891 (119910 (119905)) 119889119905 119909 isin (119886 119887) (2)

where 120582 is a complex number the kernel 119870(119909 119905) 119891(119910) and119892(119909) are known functions and 119910(119909) is an unknown functionto be determined

This paper has six sections In Section 2 a univariatetrigonometric cardinal spline on a small compact supportis constructed and properties are studied In Section 3 theapplications of trigonometric cardinal splines on solving theVolterra integral equations are explored The unknown func-tion is expressed as a linear combination of trigonometric

Hindawi Publishing CorporationISRN Applied MathematicsVolume 2014 Article ID 213909 7 pageshttpdxdoiorg1011552014213909

2 ISRN Applied Mathematics

cardinal spline functions Then a simple system of nonlinearequations on the coefficients is deducted It is relatively simpleto solve the linear system since the system is triangular anda good approximation of the original solution is obtainedThe sufficient condition for the existence is discussed and theconvergence rate is investigated In Section 4 the applicationsof trigonometric cardinal splines on solving the systems ofVolterra integral equations are explored In Section 5 numer-ical examples are given on solving the nonlinear Volterraintegral equations and a system of nonlinear Volterra integralequations Section 6 contains the conclusion remarks

2 A Cardinal Trigonometric Spline witha Small Support

To construct cardinal trigonometric splines let

120583ℎ(119909) =

1 minusℎ2lt 119909 le ℎ

2

0 elsewhere(3)

This is the zero degree polynomial or trigonometric B-spline

Let 1198790ℎ(119909) = 120583

ℎ(119909) A continuous univariate cardinal

trigonometric spline with a small support is

1198791ℎ(119909) =

11986811198790(119909)

11986811198790(0)

= 12 sin (ℎ2)

intℎ2

minusℎ2

120583ℎ(119909 + 119905) cos 119905 119889119905

(4)

Explicitly

1198791ℎ(119909) =

12 sin (ℎ2)

(sin(ℎ2) minus sin(119909 minus ℎ

2))

for 0 le 119909 le ℎ1

2 sin (ℎ2)(sin(ℎ

2) + sin(119909 + ℎ

2))

for minus ℎ le 119909 le 00 for 119909 gt ℎ or 119909 lt minusℎ

(5)

The graph of 1198791ℎ(119909) is Figure 1

Proposition 1 If 119910(119909) isin 1198621[119886 119887]11991010158401015840(119909) exists and is boundedon the finite interval [119886 119887] (where 119886 lt 119887) for any 119909 isin [119886 119887] andany integer 119899 such that ℎ = (119887 minus 119886)119899 lt 1 let

119879119871119910 (119909) =119899

sum119895=0

119910 (119886 + 119895ℎ) 1198791ℎ(119909 minus 119886 minus 119895ℎ) (6)

then |119910(119909) minus 119879119871119910((119909))| le 6ℎ2Max119909isin[119886119887]

|11991010158401015840(119909)|If 119910(119909) isin 1198621(minusinfininfin) 11991010158401015840(119909) exists on (minusinfininfin) and both

1199101015840(119909) and 11991010158401015840(119909) are bounded for any 119909 isin (minusinfininfin) and anychosen ℎ lt 1 let

119879119871 (119910 (119909)) =infin

sum119895=minusinfin

119910 (119895ℎ) 1198791ℎ(119909 minus 119895ℎ) (7)

then |119910(119909) minus 119879119871119910((119909))| le 6ℎ2Max119909isin(minusinfininfin)

|11991010158401015840(119909)|

x

y

minush h

1

Figure 1 The graph of 1198791ℎ(119909)

3 Numerical Method SolvingIntegral Equations

To solve the Volterra integral equations (2) in an interval(119886 119887) we let ℎ = (119887 minus 119886)119899 119909

119894= 119886 + 119894ℎ 119894 = 0 1 119899

Furthermore let

119910 (119909) =119899

sum119896=0

1198881198961198791ℎ(119909 minus 119909

119896)

119891 (119910 (119909)) =119899

sum119896=0

119891 (119888119896) 1198791ℎ(119909 minus 119909

119896)

119870 (119909 119905) =119899

sum119894=0

119899

sum119895=0

119870(119909119894 119909119895) 1198791ℎ(119909 minus 119909

119894) 1198791ℎ(119905 minus 119909

119895)

119892 (119909) =119899

sum119896=0

119892 (119909119896) 1198791ℎ(119909 minus 119909

119896)

(8)

plugging in (2) we get119899

sum119896=0

1198881198961198791ℎ(119909 minus 119909

119896)

minus 120582119899

sum119894=0

119899

sum119895=0

119899

sum119896=0

1198791ℎ(119909 minus 119909

119894) int119909

119886

119870(119909119894 119909119895) 1198791ℎ(119905 minus 119909

119895)

119891 (119888119896) 1198791ℎ(119905 minus 119909

119896) 119889119905

=119899

sum119896=0

119892 (119909119896) 1198791ℎ(119909 minus 119909

119896)

(9)

Letting 119909 = 1199091198960 we arrive for 119896

0= 0 1 2 3 4 119899

(where 120572 = intinfinminusinfin

11987911(119909)11987911(119909)119889119909 120573 = intinfin

minusinfin

11987911(119909)11987911(119909 minus

1)119889119909) at

119892 (1199091198960) = 1198881198960minus 120582(120572

2119891 (1198880) + 120573119891 (119888

1)) ℎ119870 (119909

0 1199091198960)

minus1198960minus1

sum119894=1

120582 (120573119891 (119888119894minus1) + 120572119891 (119888

119894)+120573119891 (119888

119894+1)) ℎ119870 (119909

119894 1199091198960)

minus 120582 (120573119891 (1198881198960minus1

) + 1205722119891 (1198881198960)) ℎ119870 (119909

1198960 1199091198960)

(1198781)

ISRN Applied Mathematics 3

which is a triangular system of 119899 + 1 nonlinear equations onunknowns 119888

0 1198881 119888

119899 Notice that the coefficient matrix for

the system is a triangular matrix which means that we solve119888119894= 119891(119888

119894) + 119882

119894 where 119882

119894is a number not depending on

119888119894 for 119894 = 0 1 119899 For the convergence rate of solution

of the Volterra integral equations (2) we have the followingProposition 2

Proposition 2 Given that 119910(119909) 119892(119909) isin 119862[119886 119887] 11991010158401015840(119909) and11989210158401015840(119909) exist and are bounded in [119886 119887]119870(119909 119910) isin 119862[119886 119887]times[119886 119887](1205972120597119909119904120597119910119905)119870(119909 119910) (119904+119905 = 2) exist and are bounded in [119886 119887]times[119886 119887] Furthermore 119870(119909 119910) satisfies the condition

100381610038161003816100381610038161003816100381610038161003816120582 int119887

119886

119870(119909 119910) (119910 (119909) minus 119906 (119909)) 119889119909100381610038161003816100381610038161003816100381610038161003816lt 119871119872max

119909isin[119886119887]

1003816100381610038161003816119910 (119909) minus 119906 (119909)1003816100381610038161003816

(10)

where |119871119872| lt 1 Let 119899 be an integer ℎ = (119887 minus 119886)119899 let 119909119894=

119886 + 119894ℎ 119910119894= 119910(119909

119894) 119894 = 0 1 2 119899 and 119888

0 1198881 119888

119899satisfies

the linear system (1198781)

119910lowast (119909) =119899

sum119896=0

1198881198941198791ℎ(119909 minus 119896ℎ) (11)

then

1003817100381710038171003817119910lowast (119909) minus 119910 (119909)1003817100381710038171003817[119886119887] = 119874 (ℎ2) (12)

where 119910(119909) is the exact solution of (2)

Proof Let

119910 (119909) =119899

sum119896=0

1198881198961198791ℎ(119909 minus 119909

119896)

119891 (119910 (119905)) =119899

sum119896=0

119891 (119888119896) 1198791ℎ(119909 minus 119909

119896)

119870 (119909 119905) =119899

sum119894=0

119899

sum119895=0

119870(119909119894 119909119895) 1198791ℎ(119909 minus 119909

119894) 1198791ℎ(119905 minus 119909

119895)

119892 (119909) =119899

sum119896=0

119892 (119909119896) 1198791ℎ(119909 minus 119909

119896)

(13)

where the coefficients are the solutions of above system (1198781)Then

1003817100381710038171003817119891 (119909) minus 119891lowast (119909)1003817100381710038171003817[119886119887]

=100381710038171003817100381710038171003817100381710038171003817120582int119887

119886

119870 (119909 119905) 119891 (119905) 119889119905 + 119892 (119909) minus 120582int119887

119886

119870lowast (119909 119905) 119891lowast (119905) 119889119905

minus119892lowast (119909)100381710038171003817100381710038171003817100381710038171003817[119886119887]

=100381710038171003817100381710038171003817100381710038171003817120582int119887

119886

119870 (119909 119905) 119891 (119905) 119889119905 minus 120582int119887

119886

119870lowast (119909 119905) 119891 (119905) 119889119905

+ 120582int119887

119886

119870lowast (119909 119905) 119891 (119905) 119889119905 minus 120582int119887

119886

119870lowast (119909 119905) 119891lowast (119905) 119889119905

+ 119892 (119909) minus 119892lowast (119909)100381710038171003817100381710038171003817100381710038171003817[119886119887]

=100381710038171003817100381710038171003817100381710038171003817120582int119887

119886

(119870 (119909 119905) minus 119870lowast (119909 119905)) 119891 (119905) 119889119905

+120582int119887

119886

119870lowast (119909 119905) (119891 (119905)minus119891lowast (119905)) 119889119905+(119892 (119909)minus119892lowast (119909))100381710038171003817100381710038171003817100381710038171003817[119886119887]

le 48 |120582|max119904+119905=2

100381710038171003817100381710038171003817100381710038171003817

1205972

120597119909119904120597119910119905119870(119909 119910)

100381710038171003817100381710038171003817100381710038171003817[119886119887]times[119886119887]

times 1003817100381710038171003817119891 (119909)1003817100381710038171003817[119886119887] (119887 minus 119886) ℎ

2

+ 1198711198721003817100381710038171003817119891 (119909) minus 119891lowast (119909)1003817100381710038171003817[119886119887] +

1003817100381710038171003817119892 (119909) minus 119892lowast (119909)1003817100381710038171003817[119886119887]

(14)Plug in

1003817100381710038171003817119891 (119909)1003817100381710038171003817[119886119887] le

11 minus 119871119872

1003817100381710038171003817119892 (119909)1003817100381710038171003817[119886119887] (15)

Therefore1003817100381710038171003817119891 (119909) minus 119891

lowast (119909)1003817100381710038171003817[119886119887]

le 11 minus 119871119872

(48 |120582| (119887 minus 119886)1 minus 119871119872

timesmax119904+119905=2

100381710038171003817100381710038171003817100381710038171003817

1205972

120597119909119904120597119910119905119870(119909 119910)

100381710038171003817100381710038171003817100381710038171003817[119886119887]times[119886119887]

times 1003817100381710038171003817119892 (119909)1003817100381710038171003817[119886119887] + 7

100381610038161003816100381610038161003816100381610038161003816100381611989210158401015840 (119909)10038161003816100381610038161003816

10038161003816100381610038161003816[119886119887])ℎ2

(16)

4 Numerical Method Solving Systems ofIntegral Equations

The system of Volterra integral equations is critical to manyphysical biological and engineering models For instancefor some heat transfer problems in physics the heat equa-tions are usually replaced by a system of Volterra integralequations [8] Many well-known models for neural networksin biomathematics nuclear reactor dynamics problems andthermoelasticity problems are also based on a system ofVolterra integral equations ([9 10]) Our method could beextended to solve the system of Volterra integral equationsGiven

119910119904(119909) = 119892

119904(119909) +

119898

sum119901=0

120582119901int119909

119886

119870119901119904(119909 119905) 119891

119901119904(119910119901(119905)) 119889119905

119909 isin (119886 119887) 119904 = 1 2 119898

(17)

4 ISRN Applied Mathematics

in an interval (119886 119887) we let ℎ = (119887 minus 119886)119899 119909119894= 119886 + 119894ℎ 119894 =

0 1 119899 Furthermore let (119904 119901 = 1 2 119898)

119910119901(119909) =

119899

sum119896=0

1198881198961199011198791ℎ(119909 minus 119909

119896)

119891119901119904(119910119901(119909)) =

119899

sum119896=0

119891119901119904(119888119896119901) 1198791ℎ(119909 minus 119909

119896)

119870119901119904(119909 119905) =

119899

sum119894=0

119899

sum119895=0

119870119901119904(119909119894 119909119895) 1198791ℎ(119909 minus 119909

119894) 1198791ℎ(119905 minus 119909

119895)

119892119904(119909) =

119899

sum119896=0

119892119904(119909119896) 1198791ℎ(119909 minus 119909

119896)

(18)

plugging in (17) we get119899

sum119896=0

1198881198961199041198791ℎ(119909 minus 119909

119896) minus119898

sum119901=0

120582119901

119899

sum119894=0

119899

sum119895=0

119899

sum119896=0

1198791ℎ(119909 minus 119909

119894)

int119909

119886

119870119901119904(119909119894 119909119895) 1198791ℎ(119905minus119909119895) 119891119901119904(119888119896119901) 1198791ℎ(119905minus119909119896) 119889119905

=119899

sum119896=0

119892119904(119909119896) 1198791ℎ(119909 minus 119909

119896)

(19)

Let 119909 = 1199091198960 we arrive for 119896

0= 0 1 2 3 4 119899 (where

120572 = intinfinminusinfin

11987911(119909)11987911(119909)119889119909 120573 = intinfin

minusinfin

11987911(119909)11987911(119909 minus 1)119889119909) at

119892119904(1199091198960)

= 1198881198960 119904

minus119898

sum119901=0

120582119901(1205722119891119901119904(1198880119904) + 120573119891

119901119904(1198881119904))

times ℎ119870119901119904(1199090 1199091198960)

minus119898

sum119901=0

120582119901

1198960minus1

sum119894=1

(120573119891119901119904(119888119894minus1119904

) + 120572119891119901119904(119888119894119904) + 120573119891

119901119904(119888119894+1119904

))

times ℎ119870119901119904(119909119894 1199091198960)

minus119898

sum119901=0

120582119901(120573119891119901119904(1198881198960minus1119904

) + 1205722119891119901119904(1198881198960 119904

)) ℎ119870119901119904(1199091198960 1199091198960)

(1198782)

which is a triangular system of (119899 + 1)119898 nonlinear equationson unknowns 119888

0119904 1198881119904 119888

119899119904 Notice that the coefficient

matrix for the system is a triangularmatrix whichmeans thatwe solve 119888

119894119904= 119891119901119904(119888119894119904) + 119882

119894119904 where 119882

119894is a number for 119894 =

0 1 119899 For the convergence rate of solution of theVolterraintegral equations (2) we have the following Proposition 3

Proposition 3 Given that 119910119901(119909) 119892119901(119909) isin 119862[119886 119887] 11991010158401015840

119901(119909) and

11989210158401015840119901(119909) exists and is bounded in [119886 119887] 119870

119901119904(119909 119910) isin 119862[119886 119887] times

[119886 119887] (1205972120597119909119904120597119910119905)119870119901119904(119909 119910) (119904 + 119905 = 2) exist and are bounded

in [119886 119887]times [119886 119887] Furthermore119870119901119904(119909 119910) satisfies the condition

10038161003816100381610038161003816100381610038161003816100381610038161003816

119898

sum119901=0

120582119901int119887

119886

119870119901119904(119909 119910) (119910

119901(119909) minus 119906

119901(119909)) 119889119909

10038161003816100381610038161003816100381610038161003816100381610038161003816

lt 119871119872max119909isin[119886119887]

119898

sum119901=0

100381610038161003816100381610038161205821199011003816100381610038161003816100381610038161003816100381610038161003816119910119901 (119909) minus 119906119901 (119909)

10038161003816100381610038161003816

(20)

where |119871119872| lt 1 Let 119899 be an integer ℎ = (119887 minus 119886)119899 let 119909119894=

119886 + 119894ℎ 119910119894= 119910(119909

119894) 119894 = 0 1 2 119899 119888

0119901 1198881119901 119888

119899119901satisfies

the linear system (1198781)

119910lowast119901(119909) =

119899

sum119896=0

1198881198941199011198791ℎ(119909 minus 119896ℎ) (21)

then10038171003817100381710038171003817119910lowast

119901(119909) minus 119910

119901(119909)10038171003817100381710038171003817[119886119887] = 119874 (ℎ2) (22)

where 119910119901(119909) is the exact solution of (17)

5 Numerical Examples

Example 1 Given that 119910(119909) = 119892(119909) + int1199090

119870(119909 119905)119891(119910(119905))119889119905where 119896(119909 119905) = tan(119909 minus 119905)(radic119905 + 01)2 119891(119910) = 1199102 119892(119909) =radic119909 + 01 + ln | cos119909|

Let (119886 119887) = (0 1) 119899 = 10 ℎ = 01 119909119894= 119894ℎ 119910(119909) =

sum10119896=0

1198881198961198791ℎ(119909 minus 119909

119896) 1199102(119905) = sum10

119896=011988821198961198791ℎ(119909 minus 119909

119896) 119870(119909 119905) =

tan(119909 minus 119905)(radic119905 + 01)2 = sum10119894=0

sum10119895=0

(tan(119909119894minus 119909119895)(radic119909119895+

01)2)1198791ℎ(119909 minus 119909

119894)1198791ℎ(119905 minus 119909

119895) 119892(119909) = sum10

119896=0119892(119909119896)1198791ℎ(119909 minus 119909

119896)

Plugging into the integral equation we arrive at

1198880= 119892 (119909

0)

119892 (119909119896) = 119888119896minus (120572

211988820+ 12057311988821) ℎ119870 (119909

0 119909119896)

minus119896minus1

sum119894=1

(1205731198882119894minus1

+ 1205721198882119894+ 1205731198882119894+1) ℎ119870 (119909

119894 119909119896)

minus (1205731198882119896minus1

+ 12057221198882119896) ℎ119870 (119909

119896 119909119896) (119896=1 2 10)

(23)

The solution is [1198880 1198881 1198882 1198883 1198884 1198885 1198886 1198887 1198888 1198889 11988810]

= [04162361433 05472481952 06478042689 0732610609908073700605 08750164888 09373063268 099540578871050163033 1102269297]

Compared with the exact solution [0416227766005472135955 06477225575 07324555320 0807106781108745966692 09366600265 09944271910 104868329811] The error 119864 lt 0003 = 3ℎ3

Example 2 119906(119909) = tan119909minus(14) sin(2119909)minus(12)119909+int119909minus12

1(1+1199062(119905))119889119905 minus12 le 119909 le 12 with the exact solution 119906(119909) =tan119909

ISRN Applied Mathematics 5

We choose 119899 = 10 ℎ = 01 and 119909119896= minus(12) + 119896ℎ and

let 119906lowast(119909) = sum10119896=0

1198881198961198791ℎ(119909 minus 119909

119896) plug into (1198781) The system has

the form

1198880= 119892 (119909

0)

119892 (119909119896) = 119888119896minus (120572

21

(11988820+ 1)

+120573

(11988821+ 1)

) ℎ

minus119896minus1

sum119894=1

(120573

(1198882119894minus1

+ 1)+ 120572(1198882119894+ 1)

+120573

(1198882119894+1

+ 1)) ℎ

minus (120573

(1198882119896minus1

+ 1)+ 1205722

1(1198882119896+ 1)

) ℎ

(119896 = 1 2 10) (24)

The solution is [1198880 1198881 1198882 1198883 1198884 1198885 1198886 1198887 1198888 1198889 11988810]

= [minus05463024898 minus04228999647 minus03095798686minus02031136045 minus01009105373 minus00007472398948009943058913 0201675057 0308203618 0421598717205450804126] Comparing with the exact solutions[119906(1199090) 119906(119909

1) 119906(119909

2) 119906(119909

3) 119906(119909

4) 119906(119909

5) 119906(119909

6) 119906(119909

7)

119906(1199098) 119906(119909

9) 119906(119909

10)] = [minus05463024898 minus04227932187

minus03093362496 minus02027100355 minus01003346720 001003346720 02027100355 03093362496 0422793218705463024898] the error 119864 lt 0002

Notice that our accuracy is much better than the one inthe paper [11] although they choose ℎ = 001 in the paper

Example 3 The equation of percolation in [12]

119910 (119909) = int119909

0

119890119860(119909minus119905) (1 + (119909 minus 119905) ln119860) (119910 (119905))1119901119889119905 (25)

where 119860 gt 1 physical parameter constant 119901 gt 1 accordingto paper [13] we can obtain a unique nonnegative nontrivialsolution Let (119886 119887) = (0 1) 119899 = 10 ℎ = 01 119909

119894= 119894ℎ

119910(119909) = sum10119896=0

1198881198961198791ℎ(119909minus119909119896)119870(119909 119905) = 119890119860(119909minus119905)(1+(119909minus119905) ln119860) =

sum10119894=0

sum10119895=0

119890119860(119909119894minus119909119895)(1 + (119909119894minus 119909119895) ln119860)119879

1ℎ(119909 minus 119909

119894)1198791ℎ(119905 minus 119909

119895)

119892(119909) = 0 Let 119860 = 2 119901 = 2 plugging into the integralequation we arrive at

1198880= 0

0 = 119888119896minus (120572

211988811199010

+ 12057311988811199011

) ℎ119870 (1199090 119909119896)

minus119896minus1

sum119894=1

(1205731198881119901119894minus1

+ 1205721198881119901119894

+ 1205731198881119901119894+1

) ℎ119870 (119909119894 119909119896)

minus (1205731198881119901119896minus1

+ 12057221198881119901119896

) ℎ119870 (119909119896 119909119896)

(119896 = 1 2 10)

(26)

The result we got is [1198880 1198881 1198882 1198883 1198884 1198885 1198886 1198887 1198888 1198889 11988810] =

[0 0003026 0013703 0035223 0071853 0129210 02146450337732 0510886 0750148 1076178]

We do not have the exact solution Nevertheless comparewith 119910(119909

119896) = int1199091198960

119890119860(119909119896minus119905)(1 + (119909119896minus 119905) ln119860)(119910(119905))1119901119889119905 [119910(119909

0)

119910(1199091) 119910(119909

2) 119910(119909

3) 119910(119909

4) 119910(119909

5) 119910(119909

6) 119910(119909

7) 119910(119909

8) 119910(119909

9)

119910(11990910)] = [0 0003011 0013632 0035039 0071478 0128539

0213538 0336002 0508289 0746365 1070795]For the same integral equation let 119860 = 2 119901 =

3 we arrive at [1198880 1198881 1198882 1198883 1198884 1198885 1198886 1198887 1198888 1198889 11988810] =

[0012903 0051423 0116203 0213289 0351521 054261808017441148333 1607165 2209747]

Compare with 119910(119909119896) = int119909119896

0

119890119860(119909119896minus119905)(1 + (119909119896

minus119905) ln119860)(119910(119905))1119901119889119905 [119910(119909

0) 119910(119909

1) 119910(119909

2) 119910(119909

3) 119910(119909

4)

119910(1199095) 119910(119909

6) 119910(119909

7) 119910(119909

8) 119910(119909

9) 119910(119909

10)] = [0012840

0051156 0115595 0212177 0349700 0539832 07976661142548 1599148 2198834]

Example 4 Consider the equation 119910(119909) minus (16) int1199090

sin(119910(119905)) cos(119909 minus 119905)119889119905 = (16) cos119909 + arcsin119909 minus (16) =119892(119909)

The exact solution is 119910 = arcsin119909Applying the method on the interval [0 1] let ℎ = 110

119899 = 10 119909119894= 119894ℎ 119870(119909 119905) = cos(119909 minus 119905) = sum11

119894=minus11198791ℎ(119909 minus

119909119894) sum11119895=minus1

cos(119909119894minus119905119895)1198791ℎ(119905minus119909119895)119892(119909) = (16) cos119909+arcsin119909minus

(16) = sum11119894=minus1

119892(119909119894)1198791ℎ(119909 minus 119909

119894) and 119910(119909) = sum119899+1

119896=minus11198881198961198791ℎ(119909 minus

119909119896) plugging into the integral equation (2) we arrive at

1198880= 119892 (119909

0)

119892 (119909119896) = 119888119896minus 120582(120572

2sin (1198880) + 120573 sin (119888

1)) ℎ119870 (119909

0 119909119896)

minus119896minus1

sum119894=1

120582 (120573 sin (119888119894minus1) + 120572 sin (119888

119894)

+120573 sin (119888119894+1)) ℎ119870 (119909

119894 119909119896)

minus 120582 (120573 sin (119888119896minus1

) + 1205722sin (119888119896)) ℎ119870 (119909

119896 119909119896)

(119896 = 1 2 10)

(27)

The solution is [1198880 1198881 1198882 1198883 1198884 1198885 1198886 1198887 1198888 1198889 11988810] =

[01001681505 02013552395 03046865884 0411506057405235819595 06434770162 0775364957 092725316781119717038 1570732818]

Compared with the exact solution [119910(1199090) 119910(119909

1) 119910(119909

2)

119910(1199093) 119910(119909

4) 119910(119909

5) 119910(119909

6) 119910(119909

7) 119910(119909

8) 119910(119909

9) 119910(119909

10)]

= [01001674211 02013579207 03046926540 0411516846005235987755 06435011087 07753974965 092729521801119769515 1570796326] The error 119864 lt 000007

Example 5 Consider the equation 119910(119909) minus (18) int1199091

(1radic1199092 + 1199052)119890119910(119905)119889119905 = 119892(119909) = ln119909minus(18)119909radic2+(18)radic1199092 + 1 theexact solution is 119910(119909) = ln119909

Applying the method on the interval [1 2] let ℎ =110 119899 = 10 119909

119894= 1 + 119894ℎ 119870(119909 119905) = 1radic1199092 + 1199052 =

sum11119894=minus1

1198791ℎ(119909 minus 119909

119894) sum11119895=minus1

119870(119909119894 119909119895)1198791ℎ(119905 minus 119909

119895) 119892(119909) = ln119909 minus

(18)119909radic2 + (18)radic1199092 + 1 = sum11119894=minus1

119892(119909119894)1198791ℎ(119909 minus 119909

119894) and

6 ISRN Applied Mathematics

119910(119909) = sum119899+1119896=minus1

1198881198961198791ℎ(119909 minus 119909

119896) plugging into the integral equa-

tion we arrive at

1198880= 119892 (119909

0)

119892 (119909119896) = 119888119896minus 120582(120572

21198901198880 + 1205731198901198881) ℎ119870 (119909

0 119909119896)

minus119896minus1

sum119894=1

120582 (120573119890119888119894minus1 + 120572119890119888119894 + 120573119890119888119894+1) ℎ119870 (119909119894 119909119896)

minus 120582 (120573119890119888119896minus1 + 1205722119890119888119896) ℎ119870 (119909

119896 119909119896)

(119896 = 1 2 10)

(28)

The solution is [1198880 1198881 1198882 1198883 1198884 1198885 1198886 1198887 1198888 1198889

11988810] = [009531139 018232333 02623661939 03364740924

04054667723 04700050464 05306294034 0587787554806418545277 06931475935]

Compared with the exact solution [119910(1199090) 119910(119909

1) 119910(119909

2)

119910(1199093) 119910(119909

4) 119910(119909

5) 119910(119909

6) 119910(119909

7) 119910(119909

8) 119910(119909

9) 119910(119909

10)] =

[009531017980 01823215567 02623642644 0336472236604054651081 04700036292 05306282510 0587786664906418538861 06931471805] The error 119864 lt 0000001

Example 6 Consider a system of Volterra integral equations

119906 (119909) = sin119909 minus 119909 + int119909

0

(1199062 (119905) + V2 (119905)) 119889119905

V (119909) = cos119909 minus 12sin2119909 + int

119909

0

(119906 (119905) V (119905)) 119889119905(29)

The exact solutions are 119906(119909) = sin119909 V(119909) = cos119909In an interval (0 1) we let ℎ = 110 119909

119894= 119894ℎ 119894 =

0 1 10 Furthermore let

119906 (119909) =119899

sum119896=0

11988811989611198791ℎ(119909 minus 119909

119896)

V (119909) =119899

sum119896=0

11988811989621198791ℎ(119909 minus 119909

119896)

(119906(119909))2 =119899

sum119896=0

(1198881198961)2 1198791ℎ(119909 minus 119909

119896)

(V(119909))2 =119899

sum119896=0

(1198881198962)2 1198791ℎ(119909 minus 119909

119896)

(119906 (119909) V (119909)) =119899

sum119896=0

(11988811989611198881198962) 1198791ℎ(119909 minus 119909

119896)

119870 (119909 119905) = 1 =119899

sum119894=0

119899

sum119895=0

1198791ℎ(119909 minus 119909

119894) 1198791ℎ(119905 minus 119909

119895)

119892 (119909) = sin119909 minus 119909 =119899

sum119896=0

119892 (119909119896) 1198791ℎ(119909 minus 119909

119896)

ℎ (119909) = cos119909 minus 12sin2119909 =

119899

sum119896=0

ℎ (119909119896) 1198791ℎ(119909 minus 119909

119896)

(30)

plugging into the system we get

11988801

= 0

119892 (119909119896) = 1198881198961

minus (1205722(119888201

+ 119888202) + 120573 (1198882

11+ 119888212)) ℎ

minus119896minus1

sum119894=1

(120573 (1198882119894minus11

+ 1198882119894minus12

) + 120572 (11988821198941+ 11988821198942)

+ 120573 (1198882119894+11

+ 1198882119894+12

)) ℎ

minus (120573 (1198882119896minus11

+ 1198882119896minus12

) + 1205722(11988821198961

+ 11988821198962)) ℎ

(119896 = 1 2 10)

11988802

= 1

ℎ (119909119896) = 1198881198962

minus (1205722(1198880111988802) + 120573 (119888

1111988812)) ℎ

minus119896minus1

sum119894=1

(120573 (119888119894minus11

119888119894minus12

) + 120572 (11988811989411198881198942) + 120573 (119888

119894+11119888119894+12

)) ℎ

minus (120573 (119888119896minus11

119888119896minus12

) + 1205722(11988811989611198881198962)) ℎ

(119896 = 1 2 10) (31)

which is a nearly triangular system of (10 + 1)2 non-linear equations on unknowns [119888

01 11988811 119888

1198991] [11988802 11988812

1198881198992] we need to solve two nonlinear equations 119888

1198941=

1198911(1198881198941 1198881198942) + 119882

1198941 1198881198942

= 1198912(1198881198941 1198881198942) + 119882

1198942 where 119882

11989411198821198941

are numbers each time for 119894 = 0 1 119899Solutions are [119888

01 11988811 119888

1198991] [11988802 11988812 119888

1198992] =

[009983172966 01986591558 02954877407 0389342381604792769003 05643832656 06438004752 071672303870782409006 08405334929] [09949873745 0979999248109551843698 09207890215 08771545950 0824714253307639887345 06955813657 06201723633 05443657395]Compared with the exact solution [0 00998334166401986693307 02955202066 03894183423 0479425538605646424733 06442176872 07173560908 0783326909608414709848] [1 09950041652 0980066577809553364891 09210609940 08775825618 0825335614907648421872 06967067093 06216099682 05403023058]The error lt0005

6 Conclusions

The proposed method is a simple and effective procedurefor solving nonlinear Volterra integral equations as well as a

ISRN Applied Mathematics 7

system of nonlinear Volterra integral equationsThemethodscan be adapted easily to the Volterra integral equations of thefirst kind which have the form 119892(119909) = int

119860

119870(119909 119905)119910(119905)119889119905 Theconvergence rate could be higher if we use more complicatedorthonormal or cardinal splines Nevertheless the resultingsystem of coefficients will be more complicated nonlinearsystems which could take more time and effort to solve

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

The work was funded by the Natural Science Foundationof Anhui Province of China under Grant no 1208085MA15and the Major Project of the Nature Science Foundation ofthe Education Department Anhui Province under Grant noKJ2014ZD30

References

[1] I J Schoenberg ldquoOn trigonometric spline interpolationrdquo Jour-nal of Mathematics and Mechanics vol 13 pp 795ndash825 1964

[2] T Lyche L L Schumaker and S Stanley ldquoQuasi-interpolantsbased on trigonometric splinesrdquo Journal of ApproximationTheory vol 95 no 2 pp 280ndash309 1998

[3] T Lyche and R Winther ldquoA stable recurrence relation fortrigonometric 119861-splinesrdquo Journal of ApproximationTheory vol25 no 3 pp 266ndash279 1979

[4] A Sharma and J Tzimbalario ldquoA class of cardinal trigonometricsplinesrdquo SIAM Journal on Mathematical Analysis vol 7 no 6pp 809ndash819 1976

[5] X Liu ldquoBivariate cardinal spline functions for digital signalprocessingrdquo inTrends in ApproximationTheory K Kopotum TLyche andM Neamtu Eds pp 261ndash271 Vanderbilt UniversityNashville Tenn USA 2001

[6] X Liu ldquoUnivariate and bivariate orthonormal splines and car-dinal splines on compact supportsrdquo Journal of Computationaland Applied Mathematics vol 195 no 1-2 pp 93ndash105 2006

[7] X Liu ldquoInterpolation by cardinal trigonometric splinesrdquo Inter-national Journal of Pure and Applied Mathematics vol 40 no 1pp 115ndash122 2007

[8] R Kress Linear Integral Equations vol 82 ofAppliedMathemat-ical Sciences Springer Berlin Germany 1989

[9] B L Moiseiwitsch Integral Equations Dover New York NYUSA 2005

[10] A D Polyanin Handbook of Integral Equations CRC PressBoca Raton Fla USA 1998

[11] A Vahidian Kamyad MMehrabinezhad and J Saberi-NadjafildquoA numerical approach for solving linear and nonlinear volterraintegral equations with controlled errorrdquo IAENG InternationalJournal of Applied Mathematics vol 40 no 2 pp 69ndash75 2010

[12] J Goncerzewicz H Marcinkowska W Okrasinski and KTabisz ldquoOn the percolation of water from a cylindrical reservoirinto the surrounding soilrdquo Zastosowania Matematyki vol 16no 2 pp 249ndash261 1978

[13] D Wei ldquoUniqueness of solutions for a class of non-linearvolterra integral equations without continuityrdquo Applied Math-ematics and Mechanics vol 18 no 12 pp 1191ndash1196 1997

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Operations ResearchAdvances in

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

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The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

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Algebra

Discrete Dynamics in Nature and Society

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Decision SciencesAdvances in

Discrete MathematicsJournal of

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

2 ISRN Applied Mathematics

cardinal spline functions Then a simple system of nonlinearequations on the coefficients is deducted It is relatively simpleto solve the linear system since the system is triangular anda good approximation of the original solution is obtainedThe sufficient condition for the existence is discussed and theconvergence rate is investigated In Section 4 the applicationsof trigonometric cardinal splines on solving the systems ofVolterra integral equations are explored In Section 5 numer-ical examples are given on solving the nonlinear Volterraintegral equations and a system of nonlinear Volterra integralequations Section 6 contains the conclusion remarks

2 A Cardinal Trigonometric Spline witha Small Support

To construct cardinal trigonometric splines let

120583ℎ(119909) =

1 minusℎ2lt 119909 le ℎ

2

0 elsewhere(3)

This is the zero degree polynomial or trigonometric B-spline

Let 1198790ℎ(119909) = 120583

ℎ(119909) A continuous univariate cardinal

trigonometric spline with a small support is

1198791ℎ(119909) =

11986811198790(119909)

11986811198790(0)

= 12 sin (ℎ2)

intℎ2

minusℎ2

120583ℎ(119909 + 119905) cos 119905 119889119905

(4)

Explicitly

1198791ℎ(119909) =

12 sin (ℎ2)

(sin(ℎ2) minus sin(119909 minus ℎ

2))

for 0 le 119909 le ℎ1

2 sin (ℎ2)(sin(ℎ

2) + sin(119909 + ℎ

2))

for minus ℎ le 119909 le 00 for 119909 gt ℎ or 119909 lt minusℎ

(5)

The graph of 1198791ℎ(119909) is Figure 1

Proposition 1 If 119910(119909) isin 1198621[119886 119887]11991010158401015840(119909) exists and is boundedon the finite interval [119886 119887] (where 119886 lt 119887) for any 119909 isin [119886 119887] andany integer 119899 such that ℎ = (119887 minus 119886)119899 lt 1 let

119879119871119910 (119909) =119899

sum119895=0

119910 (119886 + 119895ℎ) 1198791ℎ(119909 minus 119886 minus 119895ℎ) (6)

then |119910(119909) minus 119879119871119910((119909))| le 6ℎ2Max119909isin[119886119887]

|11991010158401015840(119909)|If 119910(119909) isin 1198621(minusinfininfin) 11991010158401015840(119909) exists on (minusinfininfin) and both

1199101015840(119909) and 11991010158401015840(119909) are bounded for any 119909 isin (minusinfininfin) and anychosen ℎ lt 1 let

119879119871 (119910 (119909)) =infin

sum119895=minusinfin

119910 (119895ℎ) 1198791ℎ(119909 minus 119895ℎ) (7)

then |119910(119909) minus 119879119871119910((119909))| le 6ℎ2Max119909isin(minusinfininfin)

|11991010158401015840(119909)|

x

y

minush h

1

Figure 1 The graph of 1198791ℎ(119909)

3 Numerical Method SolvingIntegral Equations

To solve the Volterra integral equations (2) in an interval(119886 119887) we let ℎ = (119887 minus 119886)119899 119909

119894= 119886 + 119894ℎ 119894 = 0 1 119899

Furthermore let

119910 (119909) =119899

sum119896=0

1198881198961198791ℎ(119909 minus 119909

119896)

119891 (119910 (119909)) =119899

sum119896=0

119891 (119888119896) 1198791ℎ(119909 minus 119909

119896)

119870 (119909 119905) =119899

sum119894=0

119899

sum119895=0

119870(119909119894 119909119895) 1198791ℎ(119909 minus 119909

119894) 1198791ℎ(119905 minus 119909

119895)

119892 (119909) =119899

sum119896=0

119892 (119909119896) 1198791ℎ(119909 minus 119909

119896)

(8)

plugging in (2) we get119899

sum119896=0

1198881198961198791ℎ(119909 minus 119909

119896)

minus 120582119899

sum119894=0

119899

sum119895=0

119899

sum119896=0

1198791ℎ(119909 minus 119909

119894) int119909

119886

119870(119909119894 119909119895) 1198791ℎ(119905 minus 119909

119895)

119891 (119888119896) 1198791ℎ(119905 minus 119909

119896) 119889119905

=119899

sum119896=0

119892 (119909119896) 1198791ℎ(119909 minus 119909

119896)

(9)

Letting 119909 = 1199091198960 we arrive for 119896

0= 0 1 2 3 4 119899

(where 120572 = intinfinminusinfin

11987911(119909)11987911(119909)119889119909 120573 = intinfin

minusinfin

11987911(119909)11987911(119909 minus

1)119889119909) at

119892 (1199091198960) = 1198881198960minus 120582(120572

2119891 (1198880) + 120573119891 (119888

1)) ℎ119870 (119909

0 1199091198960)

minus1198960minus1

sum119894=1

120582 (120573119891 (119888119894minus1) + 120572119891 (119888

119894)+120573119891 (119888

119894+1)) ℎ119870 (119909

119894 1199091198960)

minus 120582 (120573119891 (1198881198960minus1

) + 1205722119891 (1198881198960)) ℎ119870 (119909

1198960 1199091198960)

(1198781)

ISRN Applied Mathematics 3

which is a triangular system of 119899 + 1 nonlinear equations onunknowns 119888

0 1198881 119888

119899 Notice that the coefficient matrix for

the system is a triangular matrix which means that we solve119888119894= 119891(119888

119894) + 119882

119894 where 119882

119894is a number not depending on

119888119894 for 119894 = 0 1 119899 For the convergence rate of solution

of the Volterra integral equations (2) we have the followingProposition 2

Proposition 2 Given that 119910(119909) 119892(119909) isin 119862[119886 119887] 11991010158401015840(119909) and11989210158401015840(119909) exist and are bounded in [119886 119887]119870(119909 119910) isin 119862[119886 119887]times[119886 119887](1205972120597119909119904120597119910119905)119870(119909 119910) (119904+119905 = 2) exist and are bounded in [119886 119887]times[119886 119887] Furthermore 119870(119909 119910) satisfies the condition

100381610038161003816100381610038161003816100381610038161003816120582 int119887

119886

119870(119909 119910) (119910 (119909) minus 119906 (119909)) 119889119909100381610038161003816100381610038161003816100381610038161003816lt 119871119872max

119909isin[119886119887]

1003816100381610038161003816119910 (119909) minus 119906 (119909)1003816100381610038161003816

(10)

where |119871119872| lt 1 Let 119899 be an integer ℎ = (119887 minus 119886)119899 let 119909119894=

119886 + 119894ℎ 119910119894= 119910(119909

119894) 119894 = 0 1 2 119899 and 119888

0 1198881 119888

119899satisfies

the linear system (1198781)

119910lowast (119909) =119899

sum119896=0

1198881198941198791ℎ(119909 minus 119896ℎ) (11)

then

1003817100381710038171003817119910lowast (119909) minus 119910 (119909)1003817100381710038171003817[119886119887] = 119874 (ℎ2) (12)

where 119910(119909) is the exact solution of (2)

Proof Let

119910 (119909) =119899

sum119896=0

1198881198961198791ℎ(119909 minus 119909

119896)

119891 (119910 (119905)) =119899

sum119896=0

119891 (119888119896) 1198791ℎ(119909 minus 119909

119896)

119870 (119909 119905) =119899

sum119894=0

119899

sum119895=0

119870(119909119894 119909119895) 1198791ℎ(119909 minus 119909

119894) 1198791ℎ(119905 minus 119909

119895)

119892 (119909) =119899

sum119896=0

119892 (119909119896) 1198791ℎ(119909 minus 119909

119896)

(13)

where the coefficients are the solutions of above system (1198781)Then

1003817100381710038171003817119891 (119909) minus 119891lowast (119909)1003817100381710038171003817[119886119887]

=100381710038171003817100381710038171003817100381710038171003817120582int119887

119886

119870 (119909 119905) 119891 (119905) 119889119905 + 119892 (119909) minus 120582int119887

119886

119870lowast (119909 119905) 119891lowast (119905) 119889119905

minus119892lowast (119909)100381710038171003817100381710038171003817100381710038171003817[119886119887]

=100381710038171003817100381710038171003817100381710038171003817120582int119887

119886

119870 (119909 119905) 119891 (119905) 119889119905 minus 120582int119887

119886

119870lowast (119909 119905) 119891 (119905) 119889119905

+ 120582int119887

119886

119870lowast (119909 119905) 119891 (119905) 119889119905 minus 120582int119887

119886

119870lowast (119909 119905) 119891lowast (119905) 119889119905

+ 119892 (119909) minus 119892lowast (119909)100381710038171003817100381710038171003817100381710038171003817[119886119887]

=100381710038171003817100381710038171003817100381710038171003817120582int119887

119886

(119870 (119909 119905) minus 119870lowast (119909 119905)) 119891 (119905) 119889119905

+120582int119887

119886

119870lowast (119909 119905) (119891 (119905)minus119891lowast (119905)) 119889119905+(119892 (119909)minus119892lowast (119909))100381710038171003817100381710038171003817100381710038171003817[119886119887]

le 48 |120582|max119904+119905=2

100381710038171003817100381710038171003817100381710038171003817

1205972

120597119909119904120597119910119905119870(119909 119910)

100381710038171003817100381710038171003817100381710038171003817[119886119887]times[119886119887]

times 1003817100381710038171003817119891 (119909)1003817100381710038171003817[119886119887] (119887 minus 119886) ℎ

2

+ 1198711198721003817100381710038171003817119891 (119909) minus 119891lowast (119909)1003817100381710038171003817[119886119887] +

1003817100381710038171003817119892 (119909) minus 119892lowast (119909)1003817100381710038171003817[119886119887]

(14)Plug in

1003817100381710038171003817119891 (119909)1003817100381710038171003817[119886119887] le

11 minus 119871119872

1003817100381710038171003817119892 (119909)1003817100381710038171003817[119886119887] (15)

Therefore1003817100381710038171003817119891 (119909) minus 119891

lowast (119909)1003817100381710038171003817[119886119887]

le 11 minus 119871119872

(48 |120582| (119887 minus 119886)1 minus 119871119872

timesmax119904+119905=2

100381710038171003817100381710038171003817100381710038171003817

1205972

120597119909119904120597119910119905119870(119909 119910)

100381710038171003817100381710038171003817100381710038171003817[119886119887]times[119886119887]

times 1003817100381710038171003817119892 (119909)1003817100381710038171003817[119886119887] + 7

100381610038161003816100381610038161003816100381610038161003816100381611989210158401015840 (119909)10038161003816100381610038161003816

10038161003816100381610038161003816[119886119887])ℎ2

(16)

4 Numerical Method Solving Systems ofIntegral Equations

The system of Volterra integral equations is critical to manyphysical biological and engineering models For instancefor some heat transfer problems in physics the heat equa-tions are usually replaced by a system of Volterra integralequations [8] Many well-known models for neural networksin biomathematics nuclear reactor dynamics problems andthermoelasticity problems are also based on a system ofVolterra integral equations ([9 10]) Our method could beextended to solve the system of Volterra integral equationsGiven

119910119904(119909) = 119892

119904(119909) +

119898

sum119901=0

120582119901int119909

119886

119870119901119904(119909 119905) 119891

119901119904(119910119901(119905)) 119889119905

119909 isin (119886 119887) 119904 = 1 2 119898

(17)

4 ISRN Applied Mathematics

in an interval (119886 119887) we let ℎ = (119887 minus 119886)119899 119909119894= 119886 + 119894ℎ 119894 =

0 1 119899 Furthermore let (119904 119901 = 1 2 119898)

119910119901(119909) =

119899

sum119896=0

1198881198961199011198791ℎ(119909 minus 119909

119896)

119891119901119904(119910119901(119909)) =

119899

sum119896=0

119891119901119904(119888119896119901) 1198791ℎ(119909 minus 119909

119896)

119870119901119904(119909 119905) =

119899

sum119894=0

119899

sum119895=0

119870119901119904(119909119894 119909119895) 1198791ℎ(119909 minus 119909

119894) 1198791ℎ(119905 minus 119909

119895)

119892119904(119909) =

119899

sum119896=0

119892119904(119909119896) 1198791ℎ(119909 minus 119909

119896)

(18)

plugging in (17) we get119899

sum119896=0

1198881198961199041198791ℎ(119909 minus 119909

119896) minus119898

sum119901=0

120582119901

119899

sum119894=0

119899

sum119895=0

119899

sum119896=0

1198791ℎ(119909 minus 119909

119894)

int119909

119886

119870119901119904(119909119894 119909119895) 1198791ℎ(119905minus119909119895) 119891119901119904(119888119896119901) 1198791ℎ(119905minus119909119896) 119889119905

=119899

sum119896=0

119892119904(119909119896) 1198791ℎ(119909 minus 119909

119896)

(19)

Let 119909 = 1199091198960 we arrive for 119896

0= 0 1 2 3 4 119899 (where

120572 = intinfinminusinfin

11987911(119909)11987911(119909)119889119909 120573 = intinfin

minusinfin

11987911(119909)11987911(119909 minus 1)119889119909) at

119892119904(1199091198960)

= 1198881198960 119904

minus119898

sum119901=0

120582119901(1205722119891119901119904(1198880119904) + 120573119891

119901119904(1198881119904))

times ℎ119870119901119904(1199090 1199091198960)

minus119898

sum119901=0

120582119901

1198960minus1

sum119894=1

(120573119891119901119904(119888119894minus1119904

) + 120572119891119901119904(119888119894119904) + 120573119891

119901119904(119888119894+1119904

))

times ℎ119870119901119904(119909119894 1199091198960)

minus119898

sum119901=0

120582119901(120573119891119901119904(1198881198960minus1119904

) + 1205722119891119901119904(1198881198960 119904

)) ℎ119870119901119904(1199091198960 1199091198960)

(1198782)

which is a triangular system of (119899 + 1)119898 nonlinear equationson unknowns 119888

0119904 1198881119904 119888

119899119904 Notice that the coefficient

matrix for the system is a triangularmatrix whichmeans thatwe solve 119888

119894119904= 119891119901119904(119888119894119904) + 119882

119894119904 where 119882

119894is a number for 119894 =

0 1 119899 For the convergence rate of solution of theVolterraintegral equations (2) we have the following Proposition 3

Proposition 3 Given that 119910119901(119909) 119892119901(119909) isin 119862[119886 119887] 11991010158401015840

119901(119909) and

11989210158401015840119901(119909) exists and is bounded in [119886 119887] 119870

119901119904(119909 119910) isin 119862[119886 119887] times

[119886 119887] (1205972120597119909119904120597119910119905)119870119901119904(119909 119910) (119904 + 119905 = 2) exist and are bounded

in [119886 119887]times [119886 119887] Furthermore119870119901119904(119909 119910) satisfies the condition

10038161003816100381610038161003816100381610038161003816100381610038161003816

119898

sum119901=0

120582119901int119887

119886

119870119901119904(119909 119910) (119910

119901(119909) minus 119906

119901(119909)) 119889119909

10038161003816100381610038161003816100381610038161003816100381610038161003816

lt 119871119872max119909isin[119886119887]

119898

sum119901=0

100381610038161003816100381610038161205821199011003816100381610038161003816100381610038161003816100381610038161003816119910119901 (119909) minus 119906119901 (119909)

10038161003816100381610038161003816

(20)

where |119871119872| lt 1 Let 119899 be an integer ℎ = (119887 minus 119886)119899 let 119909119894=

119886 + 119894ℎ 119910119894= 119910(119909

119894) 119894 = 0 1 2 119899 119888

0119901 1198881119901 119888

119899119901satisfies

the linear system (1198781)

119910lowast119901(119909) =

119899

sum119896=0

1198881198941199011198791ℎ(119909 minus 119896ℎ) (21)

then10038171003817100381710038171003817119910lowast

119901(119909) minus 119910

119901(119909)10038171003817100381710038171003817[119886119887] = 119874 (ℎ2) (22)

where 119910119901(119909) is the exact solution of (17)

5 Numerical Examples

Example 1 Given that 119910(119909) = 119892(119909) + int1199090

119870(119909 119905)119891(119910(119905))119889119905where 119896(119909 119905) = tan(119909 minus 119905)(radic119905 + 01)2 119891(119910) = 1199102 119892(119909) =radic119909 + 01 + ln | cos119909|

Let (119886 119887) = (0 1) 119899 = 10 ℎ = 01 119909119894= 119894ℎ 119910(119909) =

sum10119896=0

1198881198961198791ℎ(119909 minus 119909

119896) 1199102(119905) = sum10

119896=011988821198961198791ℎ(119909 minus 119909

119896) 119870(119909 119905) =

tan(119909 minus 119905)(radic119905 + 01)2 = sum10119894=0

sum10119895=0

(tan(119909119894minus 119909119895)(radic119909119895+

01)2)1198791ℎ(119909 minus 119909

119894)1198791ℎ(119905 minus 119909

119895) 119892(119909) = sum10

119896=0119892(119909119896)1198791ℎ(119909 minus 119909

119896)

Plugging into the integral equation we arrive at

1198880= 119892 (119909

0)

119892 (119909119896) = 119888119896minus (120572

211988820+ 12057311988821) ℎ119870 (119909

0 119909119896)

minus119896minus1

sum119894=1

(1205731198882119894minus1

+ 1205721198882119894+ 1205731198882119894+1) ℎ119870 (119909

119894 119909119896)

minus (1205731198882119896minus1

+ 12057221198882119896) ℎ119870 (119909

119896 119909119896) (119896=1 2 10)

(23)

The solution is [1198880 1198881 1198882 1198883 1198884 1198885 1198886 1198887 1198888 1198889 11988810]

= [04162361433 05472481952 06478042689 0732610609908073700605 08750164888 09373063268 099540578871050163033 1102269297]

Compared with the exact solution [0416227766005472135955 06477225575 07324555320 0807106781108745966692 09366600265 09944271910 104868329811] The error 119864 lt 0003 = 3ℎ3

Example 2 119906(119909) = tan119909minus(14) sin(2119909)minus(12)119909+int119909minus12

1(1+1199062(119905))119889119905 minus12 le 119909 le 12 with the exact solution 119906(119909) =tan119909

ISRN Applied Mathematics 5

We choose 119899 = 10 ℎ = 01 and 119909119896= minus(12) + 119896ℎ and

let 119906lowast(119909) = sum10119896=0

1198881198961198791ℎ(119909 minus 119909

119896) plug into (1198781) The system has

the form

1198880= 119892 (119909

0)

119892 (119909119896) = 119888119896minus (120572

21

(11988820+ 1)

+120573

(11988821+ 1)

) ℎ

minus119896minus1

sum119894=1

(120573

(1198882119894minus1

+ 1)+ 120572(1198882119894+ 1)

+120573

(1198882119894+1

+ 1)) ℎ

minus (120573

(1198882119896minus1

+ 1)+ 1205722

1(1198882119896+ 1)

) ℎ

(119896 = 1 2 10) (24)

The solution is [1198880 1198881 1198882 1198883 1198884 1198885 1198886 1198887 1198888 1198889 11988810]

= [minus05463024898 minus04228999647 minus03095798686minus02031136045 minus01009105373 minus00007472398948009943058913 0201675057 0308203618 0421598717205450804126] Comparing with the exact solutions[119906(1199090) 119906(119909

1) 119906(119909

2) 119906(119909

3) 119906(119909

4) 119906(119909

5) 119906(119909

6) 119906(119909

7)

119906(1199098) 119906(119909

9) 119906(119909

10)] = [minus05463024898 minus04227932187

minus03093362496 minus02027100355 minus01003346720 001003346720 02027100355 03093362496 0422793218705463024898] the error 119864 lt 0002

Notice that our accuracy is much better than the one inthe paper [11] although they choose ℎ = 001 in the paper

Example 3 The equation of percolation in [12]

119910 (119909) = int119909

0

119890119860(119909minus119905) (1 + (119909 minus 119905) ln119860) (119910 (119905))1119901119889119905 (25)

where 119860 gt 1 physical parameter constant 119901 gt 1 accordingto paper [13] we can obtain a unique nonnegative nontrivialsolution Let (119886 119887) = (0 1) 119899 = 10 ℎ = 01 119909

119894= 119894ℎ

119910(119909) = sum10119896=0

1198881198961198791ℎ(119909minus119909119896)119870(119909 119905) = 119890119860(119909minus119905)(1+(119909minus119905) ln119860) =

sum10119894=0

sum10119895=0

119890119860(119909119894minus119909119895)(1 + (119909119894minus 119909119895) ln119860)119879

1ℎ(119909 minus 119909

119894)1198791ℎ(119905 minus 119909

119895)

119892(119909) = 0 Let 119860 = 2 119901 = 2 plugging into the integralequation we arrive at

1198880= 0

0 = 119888119896minus (120572

211988811199010

+ 12057311988811199011

) ℎ119870 (1199090 119909119896)

minus119896minus1

sum119894=1

(1205731198881119901119894minus1

+ 1205721198881119901119894

+ 1205731198881119901119894+1

) ℎ119870 (119909119894 119909119896)

minus (1205731198881119901119896minus1

+ 12057221198881119901119896

) ℎ119870 (119909119896 119909119896)

(119896 = 1 2 10)

(26)

The result we got is [1198880 1198881 1198882 1198883 1198884 1198885 1198886 1198887 1198888 1198889 11988810] =

[0 0003026 0013703 0035223 0071853 0129210 02146450337732 0510886 0750148 1076178]

We do not have the exact solution Nevertheless comparewith 119910(119909

119896) = int1199091198960

119890119860(119909119896minus119905)(1 + (119909119896minus 119905) ln119860)(119910(119905))1119901119889119905 [119910(119909

0)

119910(1199091) 119910(119909

2) 119910(119909

3) 119910(119909

4) 119910(119909

5) 119910(119909

6) 119910(119909

7) 119910(119909

8) 119910(119909

9)

119910(11990910)] = [0 0003011 0013632 0035039 0071478 0128539

0213538 0336002 0508289 0746365 1070795]For the same integral equation let 119860 = 2 119901 =

3 we arrive at [1198880 1198881 1198882 1198883 1198884 1198885 1198886 1198887 1198888 1198889 11988810] =

[0012903 0051423 0116203 0213289 0351521 054261808017441148333 1607165 2209747]

Compare with 119910(119909119896) = int119909119896

0

119890119860(119909119896minus119905)(1 + (119909119896

minus119905) ln119860)(119910(119905))1119901119889119905 [119910(119909

0) 119910(119909

1) 119910(119909

2) 119910(119909

3) 119910(119909

4)

119910(1199095) 119910(119909

6) 119910(119909

7) 119910(119909

8) 119910(119909

9) 119910(119909

10)] = [0012840

0051156 0115595 0212177 0349700 0539832 07976661142548 1599148 2198834]

Example 4 Consider the equation 119910(119909) minus (16) int1199090

sin(119910(119905)) cos(119909 minus 119905)119889119905 = (16) cos119909 + arcsin119909 minus (16) =119892(119909)

The exact solution is 119910 = arcsin119909Applying the method on the interval [0 1] let ℎ = 110

119899 = 10 119909119894= 119894ℎ 119870(119909 119905) = cos(119909 minus 119905) = sum11

119894=minus11198791ℎ(119909 minus

119909119894) sum11119895=minus1

cos(119909119894minus119905119895)1198791ℎ(119905minus119909119895)119892(119909) = (16) cos119909+arcsin119909minus

(16) = sum11119894=minus1

119892(119909119894)1198791ℎ(119909 minus 119909

119894) and 119910(119909) = sum119899+1

119896=minus11198881198961198791ℎ(119909 minus

119909119896) plugging into the integral equation (2) we arrive at

1198880= 119892 (119909

0)

119892 (119909119896) = 119888119896minus 120582(120572

2sin (1198880) + 120573 sin (119888

1)) ℎ119870 (119909

0 119909119896)

minus119896minus1

sum119894=1

120582 (120573 sin (119888119894minus1) + 120572 sin (119888

119894)

+120573 sin (119888119894+1)) ℎ119870 (119909

119894 119909119896)

minus 120582 (120573 sin (119888119896minus1

) + 1205722sin (119888119896)) ℎ119870 (119909

119896 119909119896)

(119896 = 1 2 10)

(27)

The solution is [1198880 1198881 1198882 1198883 1198884 1198885 1198886 1198887 1198888 1198889 11988810] =

[01001681505 02013552395 03046865884 0411506057405235819595 06434770162 0775364957 092725316781119717038 1570732818]

Compared with the exact solution [119910(1199090) 119910(119909

1) 119910(119909

2)

119910(1199093) 119910(119909

4) 119910(119909

5) 119910(119909

6) 119910(119909

7) 119910(119909

8) 119910(119909

9) 119910(119909

10)]

= [01001674211 02013579207 03046926540 0411516846005235987755 06435011087 07753974965 092729521801119769515 1570796326] The error 119864 lt 000007

Example 5 Consider the equation 119910(119909) minus (18) int1199091

(1radic1199092 + 1199052)119890119910(119905)119889119905 = 119892(119909) = ln119909minus(18)119909radic2+(18)radic1199092 + 1 theexact solution is 119910(119909) = ln119909

Applying the method on the interval [1 2] let ℎ =110 119899 = 10 119909

119894= 1 + 119894ℎ 119870(119909 119905) = 1radic1199092 + 1199052 =

sum11119894=minus1

1198791ℎ(119909 minus 119909

119894) sum11119895=minus1

119870(119909119894 119909119895)1198791ℎ(119905 minus 119909

119895) 119892(119909) = ln119909 minus

(18)119909radic2 + (18)radic1199092 + 1 = sum11119894=minus1

119892(119909119894)1198791ℎ(119909 minus 119909

119894) and

6 ISRN Applied Mathematics

119910(119909) = sum119899+1119896=minus1

1198881198961198791ℎ(119909 minus 119909

119896) plugging into the integral equa-

tion we arrive at

1198880= 119892 (119909

0)

119892 (119909119896) = 119888119896minus 120582(120572

21198901198880 + 1205731198901198881) ℎ119870 (119909

0 119909119896)

minus119896minus1

sum119894=1

120582 (120573119890119888119894minus1 + 120572119890119888119894 + 120573119890119888119894+1) ℎ119870 (119909119894 119909119896)

minus 120582 (120573119890119888119896minus1 + 1205722119890119888119896) ℎ119870 (119909

119896 119909119896)

(119896 = 1 2 10)

(28)

The solution is [1198880 1198881 1198882 1198883 1198884 1198885 1198886 1198887 1198888 1198889

11988810] = [009531139 018232333 02623661939 03364740924

04054667723 04700050464 05306294034 0587787554806418545277 06931475935]

Compared with the exact solution [119910(1199090) 119910(119909

1) 119910(119909

2)

119910(1199093) 119910(119909

4) 119910(119909

5) 119910(119909

6) 119910(119909

7) 119910(119909

8) 119910(119909

9) 119910(119909

10)] =

[009531017980 01823215567 02623642644 0336472236604054651081 04700036292 05306282510 0587786664906418538861 06931471805] The error 119864 lt 0000001

Example 6 Consider a system of Volterra integral equations

119906 (119909) = sin119909 minus 119909 + int119909

0

(1199062 (119905) + V2 (119905)) 119889119905

V (119909) = cos119909 minus 12sin2119909 + int

119909

0

(119906 (119905) V (119905)) 119889119905(29)

The exact solutions are 119906(119909) = sin119909 V(119909) = cos119909In an interval (0 1) we let ℎ = 110 119909

119894= 119894ℎ 119894 =

0 1 10 Furthermore let

119906 (119909) =119899

sum119896=0

11988811989611198791ℎ(119909 minus 119909

119896)

V (119909) =119899

sum119896=0

11988811989621198791ℎ(119909 minus 119909

119896)

(119906(119909))2 =119899

sum119896=0

(1198881198961)2 1198791ℎ(119909 minus 119909

119896)

(V(119909))2 =119899

sum119896=0

(1198881198962)2 1198791ℎ(119909 minus 119909

119896)

(119906 (119909) V (119909)) =119899

sum119896=0

(11988811989611198881198962) 1198791ℎ(119909 minus 119909

119896)

119870 (119909 119905) = 1 =119899

sum119894=0

119899

sum119895=0

1198791ℎ(119909 minus 119909

119894) 1198791ℎ(119905 minus 119909

119895)

119892 (119909) = sin119909 minus 119909 =119899

sum119896=0

119892 (119909119896) 1198791ℎ(119909 minus 119909

119896)

ℎ (119909) = cos119909 minus 12sin2119909 =

119899

sum119896=0

ℎ (119909119896) 1198791ℎ(119909 minus 119909

119896)

(30)

plugging into the system we get

11988801

= 0

119892 (119909119896) = 1198881198961

minus (1205722(119888201

+ 119888202) + 120573 (1198882

11+ 119888212)) ℎ

minus119896minus1

sum119894=1

(120573 (1198882119894minus11

+ 1198882119894minus12

) + 120572 (11988821198941+ 11988821198942)

+ 120573 (1198882119894+11

+ 1198882119894+12

)) ℎ

minus (120573 (1198882119896minus11

+ 1198882119896minus12

) + 1205722(11988821198961

+ 11988821198962)) ℎ

(119896 = 1 2 10)

11988802

= 1

ℎ (119909119896) = 1198881198962

minus (1205722(1198880111988802) + 120573 (119888

1111988812)) ℎ

minus119896minus1

sum119894=1

(120573 (119888119894minus11

119888119894minus12

) + 120572 (11988811989411198881198942) + 120573 (119888

119894+11119888119894+12

)) ℎ

minus (120573 (119888119896minus11

119888119896minus12

) + 1205722(11988811989611198881198962)) ℎ

(119896 = 1 2 10) (31)

which is a nearly triangular system of (10 + 1)2 non-linear equations on unknowns [119888

01 11988811 119888

1198991] [11988802 11988812

1198881198992] we need to solve two nonlinear equations 119888

1198941=

1198911(1198881198941 1198881198942) + 119882

1198941 1198881198942

= 1198912(1198881198941 1198881198942) + 119882

1198942 where 119882

11989411198821198941

are numbers each time for 119894 = 0 1 119899Solutions are [119888

01 11988811 119888

1198991] [11988802 11988812 119888

1198992] =

[009983172966 01986591558 02954877407 0389342381604792769003 05643832656 06438004752 071672303870782409006 08405334929] [09949873745 0979999248109551843698 09207890215 08771545950 0824714253307639887345 06955813657 06201723633 05443657395]Compared with the exact solution [0 00998334166401986693307 02955202066 03894183423 0479425538605646424733 06442176872 07173560908 0783326909608414709848] [1 09950041652 0980066577809553364891 09210609940 08775825618 0825335614907648421872 06967067093 06216099682 05403023058]The error lt0005

6 Conclusions

The proposed method is a simple and effective procedurefor solving nonlinear Volterra integral equations as well as a

ISRN Applied Mathematics 7

system of nonlinear Volterra integral equationsThemethodscan be adapted easily to the Volterra integral equations of thefirst kind which have the form 119892(119909) = int

119860

119870(119909 119905)119910(119905)119889119905 Theconvergence rate could be higher if we use more complicatedorthonormal or cardinal splines Nevertheless the resultingsystem of coefficients will be more complicated nonlinearsystems which could take more time and effort to solve

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

The work was funded by the Natural Science Foundationof Anhui Province of China under Grant no 1208085MA15and the Major Project of the Nature Science Foundation ofthe Education Department Anhui Province under Grant noKJ2014ZD30

References

[1] I J Schoenberg ldquoOn trigonometric spline interpolationrdquo Jour-nal of Mathematics and Mechanics vol 13 pp 795ndash825 1964

[2] T Lyche L L Schumaker and S Stanley ldquoQuasi-interpolantsbased on trigonometric splinesrdquo Journal of ApproximationTheory vol 95 no 2 pp 280ndash309 1998

[3] T Lyche and R Winther ldquoA stable recurrence relation fortrigonometric 119861-splinesrdquo Journal of ApproximationTheory vol25 no 3 pp 266ndash279 1979

[4] A Sharma and J Tzimbalario ldquoA class of cardinal trigonometricsplinesrdquo SIAM Journal on Mathematical Analysis vol 7 no 6pp 809ndash819 1976

[5] X Liu ldquoBivariate cardinal spline functions for digital signalprocessingrdquo inTrends in ApproximationTheory K Kopotum TLyche andM Neamtu Eds pp 261ndash271 Vanderbilt UniversityNashville Tenn USA 2001

[6] X Liu ldquoUnivariate and bivariate orthonormal splines and car-dinal splines on compact supportsrdquo Journal of Computationaland Applied Mathematics vol 195 no 1-2 pp 93ndash105 2006

[7] X Liu ldquoInterpolation by cardinal trigonometric splinesrdquo Inter-national Journal of Pure and Applied Mathematics vol 40 no 1pp 115ndash122 2007

[8] R Kress Linear Integral Equations vol 82 ofAppliedMathemat-ical Sciences Springer Berlin Germany 1989

[9] B L Moiseiwitsch Integral Equations Dover New York NYUSA 2005

[10] A D Polyanin Handbook of Integral Equations CRC PressBoca Raton Fla USA 1998

[11] A Vahidian Kamyad MMehrabinezhad and J Saberi-NadjafildquoA numerical approach for solving linear and nonlinear volterraintegral equations with controlled errorrdquo IAENG InternationalJournal of Applied Mathematics vol 40 no 2 pp 69ndash75 2010

[12] J Goncerzewicz H Marcinkowska W Okrasinski and KTabisz ldquoOn the percolation of water from a cylindrical reservoirinto the surrounding soilrdquo Zastosowania Matematyki vol 16no 2 pp 249ndash261 1978

[13] D Wei ldquoUniqueness of solutions for a class of non-linearvolterra integral equations without continuityrdquo Applied Math-ematics and Mechanics vol 18 no 12 pp 1191ndash1196 1997

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

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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

ISRN Applied Mathematics 3

which is a triangular system of 119899 + 1 nonlinear equations onunknowns 119888

0 1198881 119888

119899 Notice that the coefficient matrix for

the system is a triangular matrix which means that we solve119888119894= 119891(119888

119894) + 119882

119894 where 119882

119894is a number not depending on

119888119894 for 119894 = 0 1 119899 For the convergence rate of solution

of the Volterra integral equations (2) we have the followingProposition 2

Proposition 2 Given that 119910(119909) 119892(119909) isin 119862[119886 119887] 11991010158401015840(119909) and11989210158401015840(119909) exist and are bounded in [119886 119887]119870(119909 119910) isin 119862[119886 119887]times[119886 119887](1205972120597119909119904120597119910119905)119870(119909 119910) (119904+119905 = 2) exist and are bounded in [119886 119887]times[119886 119887] Furthermore 119870(119909 119910) satisfies the condition

100381610038161003816100381610038161003816100381610038161003816120582 int119887

119886

119870(119909 119910) (119910 (119909) minus 119906 (119909)) 119889119909100381610038161003816100381610038161003816100381610038161003816lt 119871119872max

119909isin[119886119887]

1003816100381610038161003816119910 (119909) minus 119906 (119909)1003816100381610038161003816

(10)

where |119871119872| lt 1 Let 119899 be an integer ℎ = (119887 minus 119886)119899 let 119909119894=

119886 + 119894ℎ 119910119894= 119910(119909

119894) 119894 = 0 1 2 119899 and 119888

0 1198881 119888

119899satisfies

the linear system (1198781)

119910lowast (119909) =119899

sum119896=0

1198881198941198791ℎ(119909 minus 119896ℎ) (11)

then

1003817100381710038171003817119910lowast (119909) minus 119910 (119909)1003817100381710038171003817[119886119887] = 119874 (ℎ2) (12)

where 119910(119909) is the exact solution of (2)

Proof Let

119910 (119909) =119899

sum119896=0

1198881198961198791ℎ(119909 minus 119909

119896)

119891 (119910 (119905)) =119899

sum119896=0

119891 (119888119896) 1198791ℎ(119909 minus 119909

119896)

119870 (119909 119905) =119899

sum119894=0

119899

sum119895=0

119870(119909119894 119909119895) 1198791ℎ(119909 minus 119909

119894) 1198791ℎ(119905 minus 119909

119895)

119892 (119909) =119899

sum119896=0

119892 (119909119896) 1198791ℎ(119909 minus 119909

119896)

(13)

where the coefficients are the solutions of above system (1198781)Then

1003817100381710038171003817119891 (119909) minus 119891lowast (119909)1003817100381710038171003817[119886119887]

=100381710038171003817100381710038171003817100381710038171003817120582int119887

119886

119870 (119909 119905) 119891 (119905) 119889119905 + 119892 (119909) minus 120582int119887

119886

119870lowast (119909 119905) 119891lowast (119905) 119889119905

minus119892lowast (119909)100381710038171003817100381710038171003817100381710038171003817[119886119887]

=100381710038171003817100381710038171003817100381710038171003817120582int119887

119886

119870 (119909 119905) 119891 (119905) 119889119905 minus 120582int119887

119886

119870lowast (119909 119905) 119891 (119905) 119889119905

+ 120582int119887

119886

119870lowast (119909 119905) 119891 (119905) 119889119905 minus 120582int119887

119886

119870lowast (119909 119905) 119891lowast (119905) 119889119905

+ 119892 (119909) minus 119892lowast (119909)100381710038171003817100381710038171003817100381710038171003817[119886119887]

=100381710038171003817100381710038171003817100381710038171003817120582int119887

119886

(119870 (119909 119905) minus 119870lowast (119909 119905)) 119891 (119905) 119889119905

+120582int119887

119886

119870lowast (119909 119905) (119891 (119905)minus119891lowast (119905)) 119889119905+(119892 (119909)minus119892lowast (119909))100381710038171003817100381710038171003817100381710038171003817[119886119887]

le 48 |120582|max119904+119905=2

100381710038171003817100381710038171003817100381710038171003817

1205972

120597119909119904120597119910119905119870(119909 119910)

100381710038171003817100381710038171003817100381710038171003817[119886119887]times[119886119887]

times 1003817100381710038171003817119891 (119909)1003817100381710038171003817[119886119887] (119887 minus 119886) ℎ

2

+ 1198711198721003817100381710038171003817119891 (119909) minus 119891lowast (119909)1003817100381710038171003817[119886119887] +

1003817100381710038171003817119892 (119909) minus 119892lowast (119909)1003817100381710038171003817[119886119887]

(14)Plug in

1003817100381710038171003817119891 (119909)1003817100381710038171003817[119886119887] le

11 minus 119871119872

1003817100381710038171003817119892 (119909)1003817100381710038171003817[119886119887] (15)

Therefore1003817100381710038171003817119891 (119909) minus 119891

lowast (119909)1003817100381710038171003817[119886119887]

le 11 minus 119871119872

(48 |120582| (119887 minus 119886)1 minus 119871119872

timesmax119904+119905=2

100381710038171003817100381710038171003817100381710038171003817

1205972

120597119909119904120597119910119905119870(119909 119910)

100381710038171003817100381710038171003817100381710038171003817[119886119887]times[119886119887]

times 1003817100381710038171003817119892 (119909)1003817100381710038171003817[119886119887] + 7

100381610038161003816100381610038161003816100381610038161003816100381611989210158401015840 (119909)10038161003816100381610038161003816

10038161003816100381610038161003816[119886119887])ℎ2

(16)

4 Numerical Method Solving Systems ofIntegral Equations

The system of Volterra integral equations is critical to manyphysical biological and engineering models For instancefor some heat transfer problems in physics the heat equa-tions are usually replaced by a system of Volterra integralequations [8] Many well-known models for neural networksin biomathematics nuclear reactor dynamics problems andthermoelasticity problems are also based on a system ofVolterra integral equations ([9 10]) Our method could beextended to solve the system of Volterra integral equationsGiven

119910119904(119909) = 119892

119904(119909) +

119898

sum119901=0

120582119901int119909

119886

119870119901119904(119909 119905) 119891

119901119904(119910119901(119905)) 119889119905

119909 isin (119886 119887) 119904 = 1 2 119898

(17)

4 ISRN Applied Mathematics

in an interval (119886 119887) we let ℎ = (119887 minus 119886)119899 119909119894= 119886 + 119894ℎ 119894 =

0 1 119899 Furthermore let (119904 119901 = 1 2 119898)

119910119901(119909) =

119899

sum119896=0

1198881198961199011198791ℎ(119909 minus 119909

119896)

119891119901119904(119910119901(119909)) =

119899

sum119896=0

119891119901119904(119888119896119901) 1198791ℎ(119909 minus 119909

119896)

119870119901119904(119909 119905) =

119899

sum119894=0

119899

sum119895=0

119870119901119904(119909119894 119909119895) 1198791ℎ(119909 minus 119909

119894) 1198791ℎ(119905 minus 119909

119895)

119892119904(119909) =

119899

sum119896=0

119892119904(119909119896) 1198791ℎ(119909 minus 119909

119896)

(18)

plugging in (17) we get119899

sum119896=0

1198881198961199041198791ℎ(119909 minus 119909

119896) minus119898

sum119901=0

120582119901

119899

sum119894=0

119899

sum119895=0

119899

sum119896=0

1198791ℎ(119909 minus 119909

119894)

int119909

119886

119870119901119904(119909119894 119909119895) 1198791ℎ(119905minus119909119895) 119891119901119904(119888119896119901) 1198791ℎ(119905minus119909119896) 119889119905

=119899

sum119896=0

119892119904(119909119896) 1198791ℎ(119909 minus 119909

119896)

(19)

Let 119909 = 1199091198960 we arrive for 119896

0= 0 1 2 3 4 119899 (where

120572 = intinfinminusinfin

11987911(119909)11987911(119909)119889119909 120573 = intinfin

minusinfin

11987911(119909)11987911(119909 minus 1)119889119909) at

119892119904(1199091198960)

= 1198881198960 119904

minus119898

sum119901=0

120582119901(1205722119891119901119904(1198880119904) + 120573119891

119901119904(1198881119904))

times ℎ119870119901119904(1199090 1199091198960)

minus119898

sum119901=0

120582119901

1198960minus1

sum119894=1

(120573119891119901119904(119888119894minus1119904

) + 120572119891119901119904(119888119894119904) + 120573119891

119901119904(119888119894+1119904

))

times ℎ119870119901119904(119909119894 1199091198960)

minus119898

sum119901=0

120582119901(120573119891119901119904(1198881198960minus1119904

) + 1205722119891119901119904(1198881198960 119904

)) ℎ119870119901119904(1199091198960 1199091198960)

(1198782)

which is a triangular system of (119899 + 1)119898 nonlinear equationson unknowns 119888

0119904 1198881119904 119888

119899119904 Notice that the coefficient

matrix for the system is a triangularmatrix whichmeans thatwe solve 119888

119894119904= 119891119901119904(119888119894119904) + 119882

119894119904 where 119882

119894is a number for 119894 =

0 1 119899 For the convergence rate of solution of theVolterraintegral equations (2) we have the following Proposition 3

Proposition 3 Given that 119910119901(119909) 119892119901(119909) isin 119862[119886 119887] 11991010158401015840

119901(119909) and

11989210158401015840119901(119909) exists and is bounded in [119886 119887] 119870

119901119904(119909 119910) isin 119862[119886 119887] times

[119886 119887] (1205972120597119909119904120597119910119905)119870119901119904(119909 119910) (119904 + 119905 = 2) exist and are bounded

in [119886 119887]times [119886 119887] Furthermore119870119901119904(119909 119910) satisfies the condition

10038161003816100381610038161003816100381610038161003816100381610038161003816

119898

sum119901=0

120582119901int119887

119886

119870119901119904(119909 119910) (119910

119901(119909) minus 119906

119901(119909)) 119889119909

10038161003816100381610038161003816100381610038161003816100381610038161003816

lt 119871119872max119909isin[119886119887]

119898

sum119901=0

100381610038161003816100381610038161205821199011003816100381610038161003816100381610038161003816100381610038161003816119910119901 (119909) minus 119906119901 (119909)

10038161003816100381610038161003816

(20)

where |119871119872| lt 1 Let 119899 be an integer ℎ = (119887 minus 119886)119899 let 119909119894=

119886 + 119894ℎ 119910119894= 119910(119909

119894) 119894 = 0 1 2 119899 119888

0119901 1198881119901 119888

119899119901satisfies

the linear system (1198781)

119910lowast119901(119909) =

119899

sum119896=0

1198881198941199011198791ℎ(119909 minus 119896ℎ) (21)

then10038171003817100381710038171003817119910lowast

119901(119909) minus 119910

119901(119909)10038171003817100381710038171003817[119886119887] = 119874 (ℎ2) (22)

where 119910119901(119909) is the exact solution of (17)

5 Numerical Examples

Example 1 Given that 119910(119909) = 119892(119909) + int1199090

119870(119909 119905)119891(119910(119905))119889119905where 119896(119909 119905) = tan(119909 minus 119905)(radic119905 + 01)2 119891(119910) = 1199102 119892(119909) =radic119909 + 01 + ln | cos119909|

Let (119886 119887) = (0 1) 119899 = 10 ℎ = 01 119909119894= 119894ℎ 119910(119909) =

sum10119896=0

1198881198961198791ℎ(119909 minus 119909

119896) 1199102(119905) = sum10

119896=011988821198961198791ℎ(119909 minus 119909

119896) 119870(119909 119905) =

tan(119909 minus 119905)(radic119905 + 01)2 = sum10119894=0

sum10119895=0

(tan(119909119894minus 119909119895)(radic119909119895+

01)2)1198791ℎ(119909 minus 119909

119894)1198791ℎ(119905 minus 119909

119895) 119892(119909) = sum10

119896=0119892(119909119896)1198791ℎ(119909 minus 119909

119896)

Plugging into the integral equation we arrive at

1198880= 119892 (119909

0)

119892 (119909119896) = 119888119896minus (120572

211988820+ 12057311988821) ℎ119870 (119909

0 119909119896)

minus119896minus1

sum119894=1

(1205731198882119894minus1

+ 1205721198882119894+ 1205731198882119894+1) ℎ119870 (119909

119894 119909119896)

minus (1205731198882119896minus1

+ 12057221198882119896) ℎ119870 (119909

119896 119909119896) (119896=1 2 10)

(23)

The solution is [1198880 1198881 1198882 1198883 1198884 1198885 1198886 1198887 1198888 1198889 11988810]

= [04162361433 05472481952 06478042689 0732610609908073700605 08750164888 09373063268 099540578871050163033 1102269297]

Compared with the exact solution [0416227766005472135955 06477225575 07324555320 0807106781108745966692 09366600265 09944271910 104868329811] The error 119864 lt 0003 = 3ℎ3

Example 2 119906(119909) = tan119909minus(14) sin(2119909)minus(12)119909+int119909minus12

1(1+1199062(119905))119889119905 minus12 le 119909 le 12 with the exact solution 119906(119909) =tan119909

ISRN Applied Mathematics 5

We choose 119899 = 10 ℎ = 01 and 119909119896= minus(12) + 119896ℎ and

let 119906lowast(119909) = sum10119896=0

1198881198961198791ℎ(119909 minus 119909

119896) plug into (1198781) The system has

the form

1198880= 119892 (119909

0)

119892 (119909119896) = 119888119896minus (120572

21

(11988820+ 1)

+120573

(11988821+ 1)

) ℎ

minus119896minus1

sum119894=1

(120573

(1198882119894minus1

+ 1)+ 120572(1198882119894+ 1)

+120573

(1198882119894+1

+ 1)) ℎ

minus (120573

(1198882119896minus1

+ 1)+ 1205722

1(1198882119896+ 1)

) ℎ

(119896 = 1 2 10) (24)

The solution is [1198880 1198881 1198882 1198883 1198884 1198885 1198886 1198887 1198888 1198889 11988810]

= [minus05463024898 minus04228999647 minus03095798686minus02031136045 minus01009105373 minus00007472398948009943058913 0201675057 0308203618 0421598717205450804126] Comparing with the exact solutions[119906(1199090) 119906(119909

1) 119906(119909

2) 119906(119909

3) 119906(119909

4) 119906(119909

5) 119906(119909

6) 119906(119909

7)

119906(1199098) 119906(119909

9) 119906(119909

10)] = [minus05463024898 minus04227932187

minus03093362496 minus02027100355 minus01003346720 001003346720 02027100355 03093362496 0422793218705463024898] the error 119864 lt 0002

Notice that our accuracy is much better than the one inthe paper [11] although they choose ℎ = 001 in the paper

Example 3 The equation of percolation in [12]

119910 (119909) = int119909

0

119890119860(119909minus119905) (1 + (119909 minus 119905) ln119860) (119910 (119905))1119901119889119905 (25)

where 119860 gt 1 physical parameter constant 119901 gt 1 accordingto paper [13] we can obtain a unique nonnegative nontrivialsolution Let (119886 119887) = (0 1) 119899 = 10 ℎ = 01 119909

119894= 119894ℎ

119910(119909) = sum10119896=0

1198881198961198791ℎ(119909minus119909119896)119870(119909 119905) = 119890119860(119909minus119905)(1+(119909minus119905) ln119860) =

sum10119894=0

sum10119895=0

119890119860(119909119894minus119909119895)(1 + (119909119894minus 119909119895) ln119860)119879

1ℎ(119909 minus 119909

119894)1198791ℎ(119905 minus 119909

119895)

119892(119909) = 0 Let 119860 = 2 119901 = 2 plugging into the integralequation we arrive at

1198880= 0

0 = 119888119896minus (120572

211988811199010

+ 12057311988811199011

) ℎ119870 (1199090 119909119896)

minus119896minus1

sum119894=1

(1205731198881119901119894minus1

+ 1205721198881119901119894

+ 1205731198881119901119894+1

) ℎ119870 (119909119894 119909119896)

minus (1205731198881119901119896minus1

+ 12057221198881119901119896

) ℎ119870 (119909119896 119909119896)

(119896 = 1 2 10)

(26)

The result we got is [1198880 1198881 1198882 1198883 1198884 1198885 1198886 1198887 1198888 1198889 11988810] =

[0 0003026 0013703 0035223 0071853 0129210 02146450337732 0510886 0750148 1076178]

We do not have the exact solution Nevertheless comparewith 119910(119909

119896) = int1199091198960

119890119860(119909119896minus119905)(1 + (119909119896minus 119905) ln119860)(119910(119905))1119901119889119905 [119910(119909

0)

119910(1199091) 119910(119909

2) 119910(119909

3) 119910(119909

4) 119910(119909

5) 119910(119909

6) 119910(119909

7) 119910(119909

8) 119910(119909

9)

119910(11990910)] = [0 0003011 0013632 0035039 0071478 0128539

0213538 0336002 0508289 0746365 1070795]For the same integral equation let 119860 = 2 119901 =

3 we arrive at [1198880 1198881 1198882 1198883 1198884 1198885 1198886 1198887 1198888 1198889 11988810] =

[0012903 0051423 0116203 0213289 0351521 054261808017441148333 1607165 2209747]

Compare with 119910(119909119896) = int119909119896

0

119890119860(119909119896minus119905)(1 + (119909119896

minus119905) ln119860)(119910(119905))1119901119889119905 [119910(119909

0) 119910(119909

1) 119910(119909

2) 119910(119909

3) 119910(119909

4)

119910(1199095) 119910(119909

6) 119910(119909

7) 119910(119909

8) 119910(119909

9) 119910(119909

10)] = [0012840

0051156 0115595 0212177 0349700 0539832 07976661142548 1599148 2198834]

Example 4 Consider the equation 119910(119909) minus (16) int1199090

sin(119910(119905)) cos(119909 minus 119905)119889119905 = (16) cos119909 + arcsin119909 minus (16) =119892(119909)

The exact solution is 119910 = arcsin119909Applying the method on the interval [0 1] let ℎ = 110

119899 = 10 119909119894= 119894ℎ 119870(119909 119905) = cos(119909 minus 119905) = sum11

119894=minus11198791ℎ(119909 minus

119909119894) sum11119895=minus1

cos(119909119894minus119905119895)1198791ℎ(119905minus119909119895)119892(119909) = (16) cos119909+arcsin119909minus

(16) = sum11119894=minus1

119892(119909119894)1198791ℎ(119909 minus 119909

119894) and 119910(119909) = sum119899+1

119896=minus11198881198961198791ℎ(119909 minus

119909119896) plugging into the integral equation (2) we arrive at

1198880= 119892 (119909

0)

119892 (119909119896) = 119888119896minus 120582(120572

2sin (1198880) + 120573 sin (119888

1)) ℎ119870 (119909

0 119909119896)

minus119896minus1

sum119894=1

120582 (120573 sin (119888119894minus1) + 120572 sin (119888

119894)

+120573 sin (119888119894+1)) ℎ119870 (119909

119894 119909119896)

minus 120582 (120573 sin (119888119896minus1

) + 1205722sin (119888119896)) ℎ119870 (119909

119896 119909119896)

(119896 = 1 2 10)

(27)

The solution is [1198880 1198881 1198882 1198883 1198884 1198885 1198886 1198887 1198888 1198889 11988810] =

[01001681505 02013552395 03046865884 0411506057405235819595 06434770162 0775364957 092725316781119717038 1570732818]

Compared with the exact solution [119910(1199090) 119910(119909

1) 119910(119909

2)

119910(1199093) 119910(119909

4) 119910(119909

5) 119910(119909

6) 119910(119909

7) 119910(119909

8) 119910(119909

9) 119910(119909

10)]

= [01001674211 02013579207 03046926540 0411516846005235987755 06435011087 07753974965 092729521801119769515 1570796326] The error 119864 lt 000007

Example 5 Consider the equation 119910(119909) minus (18) int1199091

(1radic1199092 + 1199052)119890119910(119905)119889119905 = 119892(119909) = ln119909minus(18)119909radic2+(18)radic1199092 + 1 theexact solution is 119910(119909) = ln119909

Applying the method on the interval [1 2] let ℎ =110 119899 = 10 119909

119894= 1 + 119894ℎ 119870(119909 119905) = 1radic1199092 + 1199052 =

sum11119894=minus1

1198791ℎ(119909 minus 119909

119894) sum11119895=minus1

119870(119909119894 119909119895)1198791ℎ(119905 minus 119909

119895) 119892(119909) = ln119909 minus

(18)119909radic2 + (18)radic1199092 + 1 = sum11119894=minus1

119892(119909119894)1198791ℎ(119909 minus 119909

119894) and

6 ISRN Applied Mathematics

119910(119909) = sum119899+1119896=minus1

1198881198961198791ℎ(119909 minus 119909

119896) plugging into the integral equa-

tion we arrive at

1198880= 119892 (119909

0)

119892 (119909119896) = 119888119896minus 120582(120572

21198901198880 + 1205731198901198881) ℎ119870 (119909

0 119909119896)

minus119896minus1

sum119894=1

120582 (120573119890119888119894minus1 + 120572119890119888119894 + 120573119890119888119894+1) ℎ119870 (119909119894 119909119896)

minus 120582 (120573119890119888119896minus1 + 1205722119890119888119896) ℎ119870 (119909

119896 119909119896)

(119896 = 1 2 10)

(28)

The solution is [1198880 1198881 1198882 1198883 1198884 1198885 1198886 1198887 1198888 1198889

11988810] = [009531139 018232333 02623661939 03364740924

04054667723 04700050464 05306294034 0587787554806418545277 06931475935]

Compared with the exact solution [119910(1199090) 119910(119909

1) 119910(119909

2)

119910(1199093) 119910(119909

4) 119910(119909

5) 119910(119909

6) 119910(119909

7) 119910(119909

8) 119910(119909

9) 119910(119909

10)] =

[009531017980 01823215567 02623642644 0336472236604054651081 04700036292 05306282510 0587786664906418538861 06931471805] The error 119864 lt 0000001

Example 6 Consider a system of Volterra integral equations

119906 (119909) = sin119909 minus 119909 + int119909

0

(1199062 (119905) + V2 (119905)) 119889119905

V (119909) = cos119909 minus 12sin2119909 + int

119909

0

(119906 (119905) V (119905)) 119889119905(29)

The exact solutions are 119906(119909) = sin119909 V(119909) = cos119909In an interval (0 1) we let ℎ = 110 119909

119894= 119894ℎ 119894 =

0 1 10 Furthermore let

119906 (119909) =119899

sum119896=0

11988811989611198791ℎ(119909 minus 119909

119896)

V (119909) =119899

sum119896=0

11988811989621198791ℎ(119909 minus 119909

119896)

(119906(119909))2 =119899

sum119896=0

(1198881198961)2 1198791ℎ(119909 minus 119909

119896)

(V(119909))2 =119899

sum119896=0

(1198881198962)2 1198791ℎ(119909 minus 119909

119896)

(119906 (119909) V (119909)) =119899

sum119896=0

(11988811989611198881198962) 1198791ℎ(119909 minus 119909

119896)

119870 (119909 119905) = 1 =119899

sum119894=0

119899

sum119895=0

1198791ℎ(119909 minus 119909

119894) 1198791ℎ(119905 minus 119909

119895)

119892 (119909) = sin119909 minus 119909 =119899

sum119896=0

119892 (119909119896) 1198791ℎ(119909 minus 119909

119896)

ℎ (119909) = cos119909 minus 12sin2119909 =

119899

sum119896=0

ℎ (119909119896) 1198791ℎ(119909 minus 119909

119896)

(30)

plugging into the system we get

11988801

= 0

119892 (119909119896) = 1198881198961

minus (1205722(119888201

+ 119888202) + 120573 (1198882

11+ 119888212)) ℎ

minus119896minus1

sum119894=1

(120573 (1198882119894minus11

+ 1198882119894minus12

) + 120572 (11988821198941+ 11988821198942)

+ 120573 (1198882119894+11

+ 1198882119894+12

)) ℎ

minus (120573 (1198882119896minus11

+ 1198882119896minus12

) + 1205722(11988821198961

+ 11988821198962)) ℎ

(119896 = 1 2 10)

11988802

= 1

ℎ (119909119896) = 1198881198962

minus (1205722(1198880111988802) + 120573 (119888

1111988812)) ℎ

minus119896minus1

sum119894=1

(120573 (119888119894minus11

119888119894minus12

) + 120572 (11988811989411198881198942) + 120573 (119888

119894+11119888119894+12

)) ℎ

minus (120573 (119888119896minus11

119888119896minus12

) + 1205722(11988811989611198881198962)) ℎ

(119896 = 1 2 10) (31)

which is a nearly triangular system of (10 + 1)2 non-linear equations on unknowns [119888

01 11988811 119888

1198991] [11988802 11988812

1198881198992] we need to solve two nonlinear equations 119888

1198941=

1198911(1198881198941 1198881198942) + 119882

1198941 1198881198942

= 1198912(1198881198941 1198881198942) + 119882

1198942 where 119882

11989411198821198941

are numbers each time for 119894 = 0 1 119899Solutions are [119888

01 11988811 119888

1198991] [11988802 11988812 119888

1198992] =

[009983172966 01986591558 02954877407 0389342381604792769003 05643832656 06438004752 071672303870782409006 08405334929] [09949873745 0979999248109551843698 09207890215 08771545950 0824714253307639887345 06955813657 06201723633 05443657395]Compared with the exact solution [0 00998334166401986693307 02955202066 03894183423 0479425538605646424733 06442176872 07173560908 0783326909608414709848] [1 09950041652 0980066577809553364891 09210609940 08775825618 0825335614907648421872 06967067093 06216099682 05403023058]The error lt0005

6 Conclusions

The proposed method is a simple and effective procedurefor solving nonlinear Volterra integral equations as well as a

ISRN Applied Mathematics 7

system of nonlinear Volterra integral equationsThemethodscan be adapted easily to the Volterra integral equations of thefirst kind which have the form 119892(119909) = int

119860

119870(119909 119905)119910(119905)119889119905 Theconvergence rate could be higher if we use more complicatedorthonormal or cardinal splines Nevertheless the resultingsystem of coefficients will be more complicated nonlinearsystems which could take more time and effort to solve

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

The work was funded by the Natural Science Foundationof Anhui Province of China under Grant no 1208085MA15and the Major Project of the Nature Science Foundation ofthe Education Department Anhui Province under Grant noKJ2014ZD30

References

[1] I J Schoenberg ldquoOn trigonometric spline interpolationrdquo Jour-nal of Mathematics and Mechanics vol 13 pp 795ndash825 1964

[2] T Lyche L L Schumaker and S Stanley ldquoQuasi-interpolantsbased on trigonometric splinesrdquo Journal of ApproximationTheory vol 95 no 2 pp 280ndash309 1998

[3] T Lyche and R Winther ldquoA stable recurrence relation fortrigonometric 119861-splinesrdquo Journal of ApproximationTheory vol25 no 3 pp 266ndash279 1979

[4] A Sharma and J Tzimbalario ldquoA class of cardinal trigonometricsplinesrdquo SIAM Journal on Mathematical Analysis vol 7 no 6pp 809ndash819 1976

[5] X Liu ldquoBivariate cardinal spline functions for digital signalprocessingrdquo inTrends in ApproximationTheory K Kopotum TLyche andM Neamtu Eds pp 261ndash271 Vanderbilt UniversityNashville Tenn USA 2001

[6] X Liu ldquoUnivariate and bivariate orthonormal splines and car-dinal splines on compact supportsrdquo Journal of Computationaland Applied Mathematics vol 195 no 1-2 pp 93ndash105 2006

[7] X Liu ldquoInterpolation by cardinal trigonometric splinesrdquo Inter-national Journal of Pure and Applied Mathematics vol 40 no 1pp 115ndash122 2007

[8] R Kress Linear Integral Equations vol 82 ofAppliedMathemat-ical Sciences Springer Berlin Germany 1989

[9] B L Moiseiwitsch Integral Equations Dover New York NYUSA 2005

[10] A D Polyanin Handbook of Integral Equations CRC PressBoca Raton Fla USA 1998

[11] A Vahidian Kamyad MMehrabinezhad and J Saberi-NadjafildquoA numerical approach for solving linear and nonlinear volterraintegral equations with controlled errorrdquo IAENG InternationalJournal of Applied Mathematics vol 40 no 2 pp 69ndash75 2010

[12] J Goncerzewicz H Marcinkowska W Okrasinski and KTabisz ldquoOn the percolation of water from a cylindrical reservoirinto the surrounding soilrdquo Zastosowania Matematyki vol 16no 2 pp 249ndash261 1978

[13] D Wei ldquoUniqueness of solutions for a class of non-linearvolterra integral equations without continuityrdquo Applied Math-ematics and Mechanics vol 18 no 12 pp 1191ndash1196 1997

Submit your manuscripts athttpwwwhindawicom

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Differential EquationsInternational Journal of

Volume 2014

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Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Mathematical PhysicsAdvances in

Complex AnalysisJournal of

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OptimizationJournal of

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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

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The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

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Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

4 ISRN Applied Mathematics

in an interval (119886 119887) we let ℎ = (119887 minus 119886)119899 119909119894= 119886 + 119894ℎ 119894 =

0 1 119899 Furthermore let (119904 119901 = 1 2 119898)

119910119901(119909) =

119899

sum119896=0

1198881198961199011198791ℎ(119909 minus 119909

119896)

119891119901119904(119910119901(119909)) =

119899

sum119896=0

119891119901119904(119888119896119901) 1198791ℎ(119909 minus 119909

119896)

119870119901119904(119909 119905) =

119899

sum119894=0

119899

sum119895=0

119870119901119904(119909119894 119909119895) 1198791ℎ(119909 minus 119909

119894) 1198791ℎ(119905 minus 119909

119895)

119892119904(119909) =

119899

sum119896=0

119892119904(119909119896) 1198791ℎ(119909 minus 119909

119896)

(18)

plugging in (17) we get119899

sum119896=0

1198881198961199041198791ℎ(119909 minus 119909

119896) minus119898

sum119901=0

120582119901

119899

sum119894=0

119899

sum119895=0

119899

sum119896=0

1198791ℎ(119909 minus 119909

119894)

int119909

119886

119870119901119904(119909119894 119909119895) 1198791ℎ(119905minus119909119895) 119891119901119904(119888119896119901) 1198791ℎ(119905minus119909119896) 119889119905

=119899

sum119896=0

119892119904(119909119896) 1198791ℎ(119909 minus 119909

119896)

(19)

Let 119909 = 1199091198960 we arrive for 119896

0= 0 1 2 3 4 119899 (where

120572 = intinfinminusinfin

11987911(119909)11987911(119909)119889119909 120573 = intinfin

minusinfin

11987911(119909)11987911(119909 minus 1)119889119909) at

119892119904(1199091198960)

= 1198881198960 119904

minus119898

sum119901=0

120582119901(1205722119891119901119904(1198880119904) + 120573119891

119901119904(1198881119904))

times ℎ119870119901119904(1199090 1199091198960)

minus119898

sum119901=0

120582119901

1198960minus1

sum119894=1

(120573119891119901119904(119888119894minus1119904

) + 120572119891119901119904(119888119894119904) + 120573119891

119901119904(119888119894+1119904

))

times ℎ119870119901119904(119909119894 1199091198960)

minus119898

sum119901=0

120582119901(120573119891119901119904(1198881198960minus1119904

) + 1205722119891119901119904(1198881198960 119904

)) ℎ119870119901119904(1199091198960 1199091198960)

(1198782)

which is a triangular system of (119899 + 1)119898 nonlinear equationson unknowns 119888

0119904 1198881119904 119888

119899119904 Notice that the coefficient

matrix for the system is a triangularmatrix whichmeans thatwe solve 119888

119894119904= 119891119901119904(119888119894119904) + 119882

119894119904 where 119882

119894is a number for 119894 =

0 1 119899 For the convergence rate of solution of theVolterraintegral equations (2) we have the following Proposition 3

Proposition 3 Given that 119910119901(119909) 119892119901(119909) isin 119862[119886 119887] 11991010158401015840

119901(119909) and

11989210158401015840119901(119909) exists and is bounded in [119886 119887] 119870

119901119904(119909 119910) isin 119862[119886 119887] times

[119886 119887] (1205972120597119909119904120597119910119905)119870119901119904(119909 119910) (119904 + 119905 = 2) exist and are bounded

in [119886 119887]times [119886 119887] Furthermore119870119901119904(119909 119910) satisfies the condition

10038161003816100381610038161003816100381610038161003816100381610038161003816

119898

sum119901=0

120582119901int119887

119886

119870119901119904(119909 119910) (119910

119901(119909) minus 119906

119901(119909)) 119889119909

10038161003816100381610038161003816100381610038161003816100381610038161003816

lt 119871119872max119909isin[119886119887]

119898

sum119901=0

100381610038161003816100381610038161205821199011003816100381610038161003816100381610038161003816100381610038161003816119910119901 (119909) minus 119906119901 (119909)

10038161003816100381610038161003816

(20)

where |119871119872| lt 1 Let 119899 be an integer ℎ = (119887 minus 119886)119899 let 119909119894=

119886 + 119894ℎ 119910119894= 119910(119909

119894) 119894 = 0 1 2 119899 119888

0119901 1198881119901 119888

119899119901satisfies

the linear system (1198781)

119910lowast119901(119909) =

119899

sum119896=0

1198881198941199011198791ℎ(119909 minus 119896ℎ) (21)

then10038171003817100381710038171003817119910lowast

119901(119909) minus 119910

119901(119909)10038171003817100381710038171003817[119886119887] = 119874 (ℎ2) (22)

where 119910119901(119909) is the exact solution of (17)

5 Numerical Examples

Example 1 Given that 119910(119909) = 119892(119909) + int1199090

119870(119909 119905)119891(119910(119905))119889119905where 119896(119909 119905) = tan(119909 minus 119905)(radic119905 + 01)2 119891(119910) = 1199102 119892(119909) =radic119909 + 01 + ln | cos119909|

Let (119886 119887) = (0 1) 119899 = 10 ℎ = 01 119909119894= 119894ℎ 119910(119909) =

sum10119896=0

1198881198961198791ℎ(119909 minus 119909

119896) 1199102(119905) = sum10

119896=011988821198961198791ℎ(119909 minus 119909

119896) 119870(119909 119905) =

tan(119909 minus 119905)(radic119905 + 01)2 = sum10119894=0

sum10119895=0

(tan(119909119894minus 119909119895)(radic119909119895+

01)2)1198791ℎ(119909 minus 119909

119894)1198791ℎ(119905 minus 119909

119895) 119892(119909) = sum10

119896=0119892(119909119896)1198791ℎ(119909 minus 119909

119896)

Plugging into the integral equation we arrive at

1198880= 119892 (119909

0)

119892 (119909119896) = 119888119896minus (120572

211988820+ 12057311988821) ℎ119870 (119909

0 119909119896)

minus119896minus1

sum119894=1

(1205731198882119894minus1

+ 1205721198882119894+ 1205731198882119894+1) ℎ119870 (119909

119894 119909119896)

minus (1205731198882119896minus1

+ 12057221198882119896) ℎ119870 (119909

119896 119909119896) (119896=1 2 10)

(23)

The solution is [1198880 1198881 1198882 1198883 1198884 1198885 1198886 1198887 1198888 1198889 11988810]

= [04162361433 05472481952 06478042689 0732610609908073700605 08750164888 09373063268 099540578871050163033 1102269297]

Compared with the exact solution [0416227766005472135955 06477225575 07324555320 0807106781108745966692 09366600265 09944271910 104868329811] The error 119864 lt 0003 = 3ℎ3

Example 2 119906(119909) = tan119909minus(14) sin(2119909)minus(12)119909+int119909minus12

1(1+1199062(119905))119889119905 minus12 le 119909 le 12 with the exact solution 119906(119909) =tan119909

ISRN Applied Mathematics 5

We choose 119899 = 10 ℎ = 01 and 119909119896= minus(12) + 119896ℎ and

let 119906lowast(119909) = sum10119896=0

1198881198961198791ℎ(119909 minus 119909

119896) plug into (1198781) The system has

the form

1198880= 119892 (119909

0)

119892 (119909119896) = 119888119896minus (120572

21

(11988820+ 1)

+120573

(11988821+ 1)

) ℎ

minus119896minus1

sum119894=1

(120573

(1198882119894minus1

+ 1)+ 120572(1198882119894+ 1)

+120573

(1198882119894+1

+ 1)) ℎ

minus (120573

(1198882119896minus1

+ 1)+ 1205722

1(1198882119896+ 1)

) ℎ

(119896 = 1 2 10) (24)

The solution is [1198880 1198881 1198882 1198883 1198884 1198885 1198886 1198887 1198888 1198889 11988810]

= [minus05463024898 minus04228999647 minus03095798686minus02031136045 minus01009105373 minus00007472398948009943058913 0201675057 0308203618 0421598717205450804126] Comparing with the exact solutions[119906(1199090) 119906(119909

1) 119906(119909

2) 119906(119909

3) 119906(119909

4) 119906(119909

5) 119906(119909

6) 119906(119909

7)

119906(1199098) 119906(119909

9) 119906(119909

10)] = [minus05463024898 minus04227932187

minus03093362496 minus02027100355 minus01003346720 001003346720 02027100355 03093362496 0422793218705463024898] the error 119864 lt 0002

Notice that our accuracy is much better than the one inthe paper [11] although they choose ℎ = 001 in the paper

Example 3 The equation of percolation in [12]

119910 (119909) = int119909

0

119890119860(119909minus119905) (1 + (119909 minus 119905) ln119860) (119910 (119905))1119901119889119905 (25)

where 119860 gt 1 physical parameter constant 119901 gt 1 accordingto paper [13] we can obtain a unique nonnegative nontrivialsolution Let (119886 119887) = (0 1) 119899 = 10 ℎ = 01 119909

119894= 119894ℎ

119910(119909) = sum10119896=0

1198881198961198791ℎ(119909minus119909119896)119870(119909 119905) = 119890119860(119909minus119905)(1+(119909minus119905) ln119860) =

sum10119894=0

sum10119895=0

119890119860(119909119894minus119909119895)(1 + (119909119894minus 119909119895) ln119860)119879

1ℎ(119909 minus 119909

119894)1198791ℎ(119905 minus 119909

119895)

119892(119909) = 0 Let 119860 = 2 119901 = 2 plugging into the integralequation we arrive at

1198880= 0

0 = 119888119896minus (120572

211988811199010

+ 12057311988811199011

) ℎ119870 (1199090 119909119896)

minus119896minus1

sum119894=1

(1205731198881119901119894minus1

+ 1205721198881119901119894

+ 1205731198881119901119894+1

) ℎ119870 (119909119894 119909119896)

minus (1205731198881119901119896minus1

+ 12057221198881119901119896

) ℎ119870 (119909119896 119909119896)

(119896 = 1 2 10)

(26)

The result we got is [1198880 1198881 1198882 1198883 1198884 1198885 1198886 1198887 1198888 1198889 11988810] =

[0 0003026 0013703 0035223 0071853 0129210 02146450337732 0510886 0750148 1076178]

We do not have the exact solution Nevertheless comparewith 119910(119909

119896) = int1199091198960

119890119860(119909119896minus119905)(1 + (119909119896minus 119905) ln119860)(119910(119905))1119901119889119905 [119910(119909

0)

119910(1199091) 119910(119909

2) 119910(119909

3) 119910(119909

4) 119910(119909

5) 119910(119909

6) 119910(119909

7) 119910(119909

8) 119910(119909

9)

119910(11990910)] = [0 0003011 0013632 0035039 0071478 0128539

0213538 0336002 0508289 0746365 1070795]For the same integral equation let 119860 = 2 119901 =

3 we arrive at [1198880 1198881 1198882 1198883 1198884 1198885 1198886 1198887 1198888 1198889 11988810] =

[0012903 0051423 0116203 0213289 0351521 054261808017441148333 1607165 2209747]

Compare with 119910(119909119896) = int119909119896

0

119890119860(119909119896minus119905)(1 + (119909119896

minus119905) ln119860)(119910(119905))1119901119889119905 [119910(119909

0) 119910(119909

1) 119910(119909

2) 119910(119909

3) 119910(119909

4)

119910(1199095) 119910(119909

6) 119910(119909

7) 119910(119909

8) 119910(119909

9) 119910(119909

10)] = [0012840

0051156 0115595 0212177 0349700 0539832 07976661142548 1599148 2198834]

Example 4 Consider the equation 119910(119909) minus (16) int1199090

sin(119910(119905)) cos(119909 minus 119905)119889119905 = (16) cos119909 + arcsin119909 minus (16) =119892(119909)

The exact solution is 119910 = arcsin119909Applying the method on the interval [0 1] let ℎ = 110

119899 = 10 119909119894= 119894ℎ 119870(119909 119905) = cos(119909 minus 119905) = sum11

119894=minus11198791ℎ(119909 minus

119909119894) sum11119895=minus1

cos(119909119894minus119905119895)1198791ℎ(119905minus119909119895)119892(119909) = (16) cos119909+arcsin119909minus

(16) = sum11119894=minus1

119892(119909119894)1198791ℎ(119909 minus 119909

119894) and 119910(119909) = sum119899+1

119896=minus11198881198961198791ℎ(119909 minus

119909119896) plugging into the integral equation (2) we arrive at

1198880= 119892 (119909

0)

119892 (119909119896) = 119888119896minus 120582(120572

2sin (1198880) + 120573 sin (119888

1)) ℎ119870 (119909

0 119909119896)

minus119896minus1

sum119894=1

120582 (120573 sin (119888119894minus1) + 120572 sin (119888

119894)

+120573 sin (119888119894+1)) ℎ119870 (119909

119894 119909119896)

minus 120582 (120573 sin (119888119896minus1

) + 1205722sin (119888119896)) ℎ119870 (119909

119896 119909119896)

(119896 = 1 2 10)

(27)

The solution is [1198880 1198881 1198882 1198883 1198884 1198885 1198886 1198887 1198888 1198889 11988810] =

[01001681505 02013552395 03046865884 0411506057405235819595 06434770162 0775364957 092725316781119717038 1570732818]

Compared with the exact solution [119910(1199090) 119910(119909

1) 119910(119909

2)

119910(1199093) 119910(119909

4) 119910(119909

5) 119910(119909

6) 119910(119909

7) 119910(119909

8) 119910(119909

9) 119910(119909

10)]

= [01001674211 02013579207 03046926540 0411516846005235987755 06435011087 07753974965 092729521801119769515 1570796326] The error 119864 lt 000007

Example 5 Consider the equation 119910(119909) minus (18) int1199091

(1radic1199092 + 1199052)119890119910(119905)119889119905 = 119892(119909) = ln119909minus(18)119909radic2+(18)radic1199092 + 1 theexact solution is 119910(119909) = ln119909

Applying the method on the interval [1 2] let ℎ =110 119899 = 10 119909

119894= 1 + 119894ℎ 119870(119909 119905) = 1radic1199092 + 1199052 =

sum11119894=minus1

1198791ℎ(119909 minus 119909

119894) sum11119895=minus1

119870(119909119894 119909119895)1198791ℎ(119905 minus 119909

119895) 119892(119909) = ln119909 minus

(18)119909radic2 + (18)radic1199092 + 1 = sum11119894=minus1

119892(119909119894)1198791ℎ(119909 minus 119909

119894) and

6 ISRN Applied Mathematics

119910(119909) = sum119899+1119896=minus1

1198881198961198791ℎ(119909 minus 119909

119896) plugging into the integral equa-

tion we arrive at

1198880= 119892 (119909

0)

119892 (119909119896) = 119888119896minus 120582(120572

21198901198880 + 1205731198901198881) ℎ119870 (119909

0 119909119896)

minus119896minus1

sum119894=1

120582 (120573119890119888119894minus1 + 120572119890119888119894 + 120573119890119888119894+1) ℎ119870 (119909119894 119909119896)

minus 120582 (120573119890119888119896minus1 + 1205722119890119888119896) ℎ119870 (119909

119896 119909119896)

(119896 = 1 2 10)

(28)

The solution is [1198880 1198881 1198882 1198883 1198884 1198885 1198886 1198887 1198888 1198889

11988810] = [009531139 018232333 02623661939 03364740924

04054667723 04700050464 05306294034 0587787554806418545277 06931475935]

Compared with the exact solution [119910(1199090) 119910(119909

1) 119910(119909

2)

119910(1199093) 119910(119909

4) 119910(119909

5) 119910(119909

6) 119910(119909

7) 119910(119909

8) 119910(119909

9) 119910(119909

10)] =

[009531017980 01823215567 02623642644 0336472236604054651081 04700036292 05306282510 0587786664906418538861 06931471805] The error 119864 lt 0000001

Example 6 Consider a system of Volterra integral equations

119906 (119909) = sin119909 minus 119909 + int119909

0

(1199062 (119905) + V2 (119905)) 119889119905

V (119909) = cos119909 minus 12sin2119909 + int

119909

0

(119906 (119905) V (119905)) 119889119905(29)

The exact solutions are 119906(119909) = sin119909 V(119909) = cos119909In an interval (0 1) we let ℎ = 110 119909

119894= 119894ℎ 119894 =

0 1 10 Furthermore let

119906 (119909) =119899

sum119896=0

11988811989611198791ℎ(119909 minus 119909

119896)

V (119909) =119899

sum119896=0

11988811989621198791ℎ(119909 minus 119909

119896)

(119906(119909))2 =119899

sum119896=0

(1198881198961)2 1198791ℎ(119909 minus 119909

119896)

(V(119909))2 =119899

sum119896=0

(1198881198962)2 1198791ℎ(119909 minus 119909

119896)

(119906 (119909) V (119909)) =119899

sum119896=0

(11988811989611198881198962) 1198791ℎ(119909 minus 119909

119896)

119870 (119909 119905) = 1 =119899

sum119894=0

119899

sum119895=0

1198791ℎ(119909 minus 119909

119894) 1198791ℎ(119905 minus 119909

119895)

119892 (119909) = sin119909 minus 119909 =119899

sum119896=0

119892 (119909119896) 1198791ℎ(119909 minus 119909

119896)

ℎ (119909) = cos119909 minus 12sin2119909 =

119899

sum119896=0

ℎ (119909119896) 1198791ℎ(119909 minus 119909

119896)

(30)

plugging into the system we get

11988801

= 0

119892 (119909119896) = 1198881198961

minus (1205722(119888201

+ 119888202) + 120573 (1198882

11+ 119888212)) ℎ

minus119896minus1

sum119894=1

(120573 (1198882119894minus11

+ 1198882119894minus12

) + 120572 (11988821198941+ 11988821198942)

+ 120573 (1198882119894+11

+ 1198882119894+12

)) ℎ

minus (120573 (1198882119896minus11

+ 1198882119896minus12

) + 1205722(11988821198961

+ 11988821198962)) ℎ

(119896 = 1 2 10)

11988802

= 1

ℎ (119909119896) = 1198881198962

minus (1205722(1198880111988802) + 120573 (119888

1111988812)) ℎ

minus119896minus1

sum119894=1

(120573 (119888119894minus11

119888119894minus12

) + 120572 (11988811989411198881198942) + 120573 (119888

119894+11119888119894+12

)) ℎ

minus (120573 (119888119896minus11

119888119896minus12

) + 1205722(11988811989611198881198962)) ℎ

(119896 = 1 2 10) (31)

which is a nearly triangular system of (10 + 1)2 non-linear equations on unknowns [119888

01 11988811 119888

1198991] [11988802 11988812

1198881198992] we need to solve two nonlinear equations 119888

1198941=

1198911(1198881198941 1198881198942) + 119882

1198941 1198881198942

= 1198912(1198881198941 1198881198942) + 119882

1198942 where 119882

11989411198821198941

are numbers each time for 119894 = 0 1 119899Solutions are [119888

01 11988811 119888

1198991] [11988802 11988812 119888

1198992] =

[009983172966 01986591558 02954877407 0389342381604792769003 05643832656 06438004752 071672303870782409006 08405334929] [09949873745 0979999248109551843698 09207890215 08771545950 0824714253307639887345 06955813657 06201723633 05443657395]Compared with the exact solution [0 00998334166401986693307 02955202066 03894183423 0479425538605646424733 06442176872 07173560908 0783326909608414709848] [1 09950041652 0980066577809553364891 09210609940 08775825618 0825335614907648421872 06967067093 06216099682 05403023058]The error lt0005

6 Conclusions

The proposed method is a simple and effective procedurefor solving nonlinear Volterra integral equations as well as a

ISRN Applied Mathematics 7

system of nonlinear Volterra integral equationsThemethodscan be adapted easily to the Volterra integral equations of thefirst kind which have the form 119892(119909) = int

119860

119870(119909 119905)119910(119905)119889119905 Theconvergence rate could be higher if we use more complicatedorthonormal or cardinal splines Nevertheless the resultingsystem of coefficients will be more complicated nonlinearsystems which could take more time and effort to solve

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

The work was funded by the Natural Science Foundationof Anhui Province of China under Grant no 1208085MA15and the Major Project of the Nature Science Foundation ofthe Education Department Anhui Province under Grant noKJ2014ZD30

References

[1] I J Schoenberg ldquoOn trigonometric spline interpolationrdquo Jour-nal of Mathematics and Mechanics vol 13 pp 795ndash825 1964

[2] T Lyche L L Schumaker and S Stanley ldquoQuasi-interpolantsbased on trigonometric splinesrdquo Journal of ApproximationTheory vol 95 no 2 pp 280ndash309 1998

[3] T Lyche and R Winther ldquoA stable recurrence relation fortrigonometric 119861-splinesrdquo Journal of ApproximationTheory vol25 no 3 pp 266ndash279 1979

[4] A Sharma and J Tzimbalario ldquoA class of cardinal trigonometricsplinesrdquo SIAM Journal on Mathematical Analysis vol 7 no 6pp 809ndash819 1976

[5] X Liu ldquoBivariate cardinal spline functions for digital signalprocessingrdquo inTrends in ApproximationTheory K Kopotum TLyche andM Neamtu Eds pp 261ndash271 Vanderbilt UniversityNashville Tenn USA 2001

[6] X Liu ldquoUnivariate and bivariate orthonormal splines and car-dinal splines on compact supportsrdquo Journal of Computationaland Applied Mathematics vol 195 no 1-2 pp 93ndash105 2006

[7] X Liu ldquoInterpolation by cardinal trigonometric splinesrdquo Inter-national Journal of Pure and Applied Mathematics vol 40 no 1pp 115ndash122 2007

[8] R Kress Linear Integral Equations vol 82 ofAppliedMathemat-ical Sciences Springer Berlin Germany 1989

[9] B L Moiseiwitsch Integral Equations Dover New York NYUSA 2005

[10] A D Polyanin Handbook of Integral Equations CRC PressBoca Raton Fla USA 1998

[11] A Vahidian Kamyad MMehrabinezhad and J Saberi-NadjafildquoA numerical approach for solving linear and nonlinear volterraintegral equations with controlled errorrdquo IAENG InternationalJournal of Applied Mathematics vol 40 no 2 pp 69ndash75 2010

[12] J Goncerzewicz H Marcinkowska W Okrasinski and KTabisz ldquoOn the percolation of water from a cylindrical reservoirinto the surrounding soilrdquo Zastosowania Matematyki vol 16no 2 pp 249ndash261 1978

[13] D Wei ldquoUniqueness of solutions for a class of non-linearvolterra integral equations without continuityrdquo Applied Math-ematics and Mechanics vol 18 no 12 pp 1191ndash1196 1997

Submit your manuscripts athttpwwwhindawicom

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Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Complex AnalysisJournal of

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OptimizationJournal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

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Operations ResearchAdvances in

Journal of

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

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Algebra

Discrete Dynamics in Nature and Society

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Decision SciencesAdvances in

Discrete MathematicsJournal of

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

ISRN Applied Mathematics 5

We choose 119899 = 10 ℎ = 01 and 119909119896= minus(12) + 119896ℎ and

let 119906lowast(119909) = sum10119896=0

1198881198961198791ℎ(119909 minus 119909

119896) plug into (1198781) The system has

the form

1198880= 119892 (119909

0)

119892 (119909119896) = 119888119896minus (120572

21

(11988820+ 1)

+120573

(11988821+ 1)

) ℎ

minus119896minus1

sum119894=1

(120573

(1198882119894minus1

+ 1)+ 120572(1198882119894+ 1)

+120573

(1198882119894+1

+ 1)) ℎ

minus (120573

(1198882119896minus1

+ 1)+ 1205722

1(1198882119896+ 1)

) ℎ

(119896 = 1 2 10) (24)

The solution is [1198880 1198881 1198882 1198883 1198884 1198885 1198886 1198887 1198888 1198889 11988810]

= [minus05463024898 minus04228999647 minus03095798686minus02031136045 minus01009105373 minus00007472398948009943058913 0201675057 0308203618 0421598717205450804126] Comparing with the exact solutions[119906(1199090) 119906(119909

1) 119906(119909

2) 119906(119909

3) 119906(119909

4) 119906(119909

5) 119906(119909

6) 119906(119909

7)

119906(1199098) 119906(119909

9) 119906(119909

10)] = [minus05463024898 minus04227932187

minus03093362496 minus02027100355 minus01003346720 001003346720 02027100355 03093362496 0422793218705463024898] the error 119864 lt 0002

Notice that our accuracy is much better than the one inthe paper [11] although they choose ℎ = 001 in the paper

Example 3 The equation of percolation in [12]

119910 (119909) = int119909

0

119890119860(119909minus119905) (1 + (119909 minus 119905) ln119860) (119910 (119905))1119901119889119905 (25)

where 119860 gt 1 physical parameter constant 119901 gt 1 accordingto paper [13] we can obtain a unique nonnegative nontrivialsolution Let (119886 119887) = (0 1) 119899 = 10 ℎ = 01 119909

119894= 119894ℎ

119910(119909) = sum10119896=0

1198881198961198791ℎ(119909minus119909119896)119870(119909 119905) = 119890119860(119909minus119905)(1+(119909minus119905) ln119860) =

sum10119894=0

sum10119895=0

119890119860(119909119894minus119909119895)(1 + (119909119894minus 119909119895) ln119860)119879

1ℎ(119909 minus 119909

119894)1198791ℎ(119905 minus 119909

119895)

119892(119909) = 0 Let 119860 = 2 119901 = 2 plugging into the integralequation we arrive at

1198880= 0

0 = 119888119896minus (120572

211988811199010

+ 12057311988811199011

) ℎ119870 (1199090 119909119896)

minus119896minus1

sum119894=1

(1205731198881119901119894minus1

+ 1205721198881119901119894

+ 1205731198881119901119894+1

) ℎ119870 (119909119894 119909119896)

minus (1205731198881119901119896minus1

+ 12057221198881119901119896

) ℎ119870 (119909119896 119909119896)

(119896 = 1 2 10)

(26)

The result we got is [1198880 1198881 1198882 1198883 1198884 1198885 1198886 1198887 1198888 1198889 11988810] =

[0 0003026 0013703 0035223 0071853 0129210 02146450337732 0510886 0750148 1076178]

We do not have the exact solution Nevertheless comparewith 119910(119909

119896) = int1199091198960

119890119860(119909119896minus119905)(1 + (119909119896minus 119905) ln119860)(119910(119905))1119901119889119905 [119910(119909

0)

119910(1199091) 119910(119909

2) 119910(119909

3) 119910(119909

4) 119910(119909

5) 119910(119909

6) 119910(119909

7) 119910(119909

8) 119910(119909

9)

119910(11990910)] = [0 0003011 0013632 0035039 0071478 0128539

0213538 0336002 0508289 0746365 1070795]For the same integral equation let 119860 = 2 119901 =

3 we arrive at [1198880 1198881 1198882 1198883 1198884 1198885 1198886 1198887 1198888 1198889 11988810] =

[0012903 0051423 0116203 0213289 0351521 054261808017441148333 1607165 2209747]

Compare with 119910(119909119896) = int119909119896

0

119890119860(119909119896minus119905)(1 + (119909119896

minus119905) ln119860)(119910(119905))1119901119889119905 [119910(119909

0) 119910(119909

1) 119910(119909

2) 119910(119909

3) 119910(119909

4)

119910(1199095) 119910(119909

6) 119910(119909

7) 119910(119909

8) 119910(119909

9) 119910(119909

10)] = [0012840

0051156 0115595 0212177 0349700 0539832 07976661142548 1599148 2198834]

Example 4 Consider the equation 119910(119909) minus (16) int1199090

sin(119910(119905)) cos(119909 minus 119905)119889119905 = (16) cos119909 + arcsin119909 minus (16) =119892(119909)

The exact solution is 119910 = arcsin119909Applying the method on the interval [0 1] let ℎ = 110

119899 = 10 119909119894= 119894ℎ 119870(119909 119905) = cos(119909 minus 119905) = sum11

119894=minus11198791ℎ(119909 minus

119909119894) sum11119895=minus1

cos(119909119894minus119905119895)1198791ℎ(119905minus119909119895)119892(119909) = (16) cos119909+arcsin119909minus

(16) = sum11119894=minus1

119892(119909119894)1198791ℎ(119909 minus 119909

119894) and 119910(119909) = sum119899+1

119896=minus11198881198961198791ℎ(119909 minus

119909119896) plugging into the integral equation (2) we arrive at

1198880= 119892 (119909

0)

119892 (119909119896) = 119888119896minus 120582(120572

2sin (1198880) + 120573 sin (119888

1)) ℎ119870 (119909

0 119909119896)

minus119896minus1

sum119894=1

120582 (120573 sin (119888119894minus1) + 120572 sin (119888

119894)

+120573 sin (119888119894+1)) ℎ119870 (119909

119894 119909119896)

minus 120582 (120573 sin (119888119896minus1

) + 1205722sin (119888119896)) ℎ119870 (119909

119896 119909119896)

(119896 = 1 2 10)

(27)

The solution is [1198880 1198881 1198882 1198883 1198884 1198885 1198886 1198887 1198888 1198889 11988810] =

[01001681505 02013552395 03046865884 0411506057405235819595 06434770162 0775364957 092725316781119717038 1570732818]

Compared with the exact solution [119910(1199090) 119910(119909

1) 119910(119909

2)

119910(1199093) 119910(119909

4) 119910(119909

5) 119910(119909

6) 119910(119909

7) 119910(119909

8) 119910(119909

9) 119910(119909

10)]

= [01001674211 02013579207 03046926540 0411516846005235987755 06435011087 07753974965 092729521801119769515 1570796326] The error 119864 lt 000007

Example 5 Consider the equation 119910(119909) minus (18) int1199091

(1radic1199092 + 1199052)119890119910(119905)119889119905 = 119892(119909) = ln119909minus(18)119909radic2+(18)radic1199092 + 1 theexact solution is 119910(119909) = ln119909

Applying the method on the interval [1 2] let ℎ =110 119899 = 10 119909

119894= 1 + 119894ℎ 119870(119909 119905) = 1radic1199092 + 1199052 =

sum11119894=minus1

1198791ℎ(119909 minus 119909

119894) sum11119895=minus1

119870(119909119894 119909119895)1198791ℎ(119905 minus 119909

119895) 119892(119909) = ln119909 minus

(18)119909radic2 + (18)radic1199092 + 1 = sum11119894=minus1

119892(119909119894)1198791ℎ(119909 minus 119909

119894) and

6 ISRN Applied Mathematics

119910(119909) = sum119899+1119896=minus1

1198881198961198791ℎ(119909 minus 119909

119896) plugging into the integral equa-

tion we arrive at

1198880= 119892 (119909

0)

119892 (119909119896) = 119888119896minus 120582(120572

21198901198880 + 1205731198901198881) ℎ119870 (119909

0 119909119896)

minus119896minus1

sum119894=1

120582 (120573119890119888119894minus1 + 120572119890119888119894 + 120573119890119888119894+1) ℎ119870 (119909119894 119909119896)

minus 120582 (120573119890119888119896minus1 + 1205722119890119888119896) ℎ119870 (119909

119896 119909119896)

(119896 = 1 2 10)

(28)

The solution is [1198880 1198881 1198882 1198883 1198884 1198885 1198886 1198887 1198888 1198889

11988810] = [009531139 018232333 02623661939 03364740924

04054667723 04700050464 05306294034 0587787554806418545277 06931475935]

Compared with the exact solution [119910(1199090) 119910(119909

1) 119910(119909

2)

119910(1199093) 119910(119909

4) 119910(119909

5) 119910(119909

6) 119910(119909

7) 119910(119909

8) 119910(119909

9) 119910(119909

10)] =

[009531017980 01823215567 02623642644 0336472236604054651081 04700036292 05306282510 0587786664906418538861 06931471805] The error 119864 lt 0000001

Example 6 Consider a system of Volterra integral equations

119906 (119909) = sin119909 minus 119909 + int119909

0

(1199062 (119905) + V2 (119905)) 119889119905

V (119909) = cos119909 minus 12sin2119909 + int

119909

0

(119906 (119905) V (119905)) 119889119905(29)

The exact solutions are 119906(119909) = sin119909 V(119909) = cos119909In an interval (0 1) we let ℎ = 110 119909

119894= 119894ℎ 119894 =

0 1 10 Furthermore let

119906 (119909) =119899

sum119896=0

11988811989611198791ℎ(119909 minus 119909

119896)

V (119909) =119899

sum119896=0

11988811989621198791ℎ(119909 minus 119909

119896)

(119906(119909))2 =119899

sum119896=0

(1198881198961)2 1198791ℎ(119909 minus 119909

119896)

(V(119909))2 =119899

sum119896=0

(1198881198962)2 1198791ℎ(119909 minus 119909

119896)

(119906 (119909) V (119909)) =119899

sum119896=0

(11988811989611198881198962) 1198791ℎ(119909 minus 119909

119896)

119870 (119909 119905) = 1 =119899

sum119894=0

119899

sum119895=0

1198791ℎ(119909 minus 119909

119894) 1198791ℎ(119905 minus 119909

119895)

119892 (119909) = sin119909 minus 119909 =119899

sum119896=0

119892 (119909119896) 1198791ℎ(119909 minus 119909

119896)

ℎ (119909) = cos119909 minus 12sin2119909 =

119899

sum119896=0

ℎ (119909119896) 1198791ℎ(119909 minus 119909

119896)

(30)

plugging into the system we get

11988801

= 0

119892 (119909119896) = 1198881198961

minus (1205722(119888201

+ 119888202) + 120573 (1198882

11+ 119888212)) ℎ

minus119896minus1

sum119894=1

(120573 (1198882119894minus11

+ 1198882119894minus12

) + 120572 (11988821198941+ 11988821198942)

+ 120573 (1198882119894+11

+ 1198882119894+12

)) ℎ

minus (120573 (1198882119896minus11

+ 1198882119896minus12

) + 1205722(11988821198961

+ 11988821198962)) ℎ

(119896 = 1 2 10)

11988802

= 1

ℎ (119909119896) = 1198881198962

minus (1205722(1198880111988802) + 120573 (119888

1111988812)) ℎ

minus119896minus1

sum119894=1

(120573 (119888119894minus11

119888119894minus12

) + 120572 (11988811989411198881198942) + 120573 (119888

119894+11119888119894+12

)) ℎ

minus (120573 (119888119896minus11

119888119896minus12

) + 1205722(11988811989611198881198962)) ℎ

(119896 = 1 2 10) (31)

which is a nearly triangular system of (10 + 1)2 non-linear equations on unknowns [119888

01 11988811 119888

1198991] [11988802 11988812

1198881198992] we need to solve two nonlinear equations 119888

1198941=

1198911(1198881198941 1198881198942) + 119882

1198941 1198881198942

= 1198912(1198881198941 1198881198942) + 119882

1198942 where 119882

11989411198821198941

are numbers each time for 119894 = 0 1 119899Solutions are [119888

01 11988811 119888

1198991] [11988802 11988812 119888

1198992] =

[009983172966 01986591558 02954877407 0389342381604792769003 05643832656 06438004752 071672303870782409006 08405334929] [09949873745 0979999248109551843698 09207890215 08771545950 0824714253307639887345 06955813657 06201723633 05443657395]Compared with the exact solution [0 00998334166401986693307 02955202066 03894183423 0479425538605646424733 06442176872 07173560908 0783326909608414709848] [1 09950041652 0980066577809553364891 09210609940 08775825618 0825335614907648421872 06967067093 06216099682 05403023058]The error lt0005

6 Conclusions

The proposed method is a simple and effective procedurefor solving nonlinear Volterra integral equations as well as a

ISRN Applied Mathematics 7

system of nonlinear Volterra integral equationsThemethodscan be adapted easily to the Volterra integral equations of thefirst kind which have the form 119892(119909) = int

119860

119870(119909 119905)119910(119905)119889119905 Theconvergence rate could be higher if we use more complicatedorthonormal or cardinal splines Nevertheless the resultingsystem of coefficients will be more complicated nonlinearsystems which could take more time and effort to solve

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

The work was funded by the Natural Science Foundationof Anhui Province of China under Grant no 1208085MA15and the Major Project of the Nature Science Foundation ofthe Education Department Anhui Province under Grant noKJ2014ZD30

References

[1] I J Schoenberg ldquoOn trigonometric spline interpolationrdquo Jour-nal of Mathematics and Mechanics vol 13 pp 795ndash825 1964

[2] T Lyche L L Schumaker and S Stanley ldquoQuasi-interpolantsbased on trigonometric splinesrdquo Journal of ApproximationTheory vol 95 no 2 pp 280ndash309 1998

[3] T Lyche and R Winther ldquoA stable recurrence relation fortrigonometric 119861-splinesrdquo Journal of ApproximationTheory vol25 no 3 pp 266ndash279 1979

[4] A Sharma and J Tzimbalario ldquoA class of cardinal trigonometricsplinesrdquo SIAM Journal on Mathematical Analysis vol 7 no 6pp 809ndash819 1976

[5] X Liu ldquoBivariate cardinal spline functions for digital signalprocessingrdquo inTrends in ApproximationTheory K Kopotum TLyche andM Neamtu Eds pp 261ndash271 Vanderbilt UniversityNashville Tenn USA 2001

[6] X Liu ldquoUnivariate and bivariate orthonormal splines and car-dinal splines on compact supportsrdquo Journal of Computationaland Applied Mathematics vol 195 no 1-2 pp 93ndash105 2006

[7] X Liu ldquoInterpolation by cardinal trigonometric splinesrdquo Inter-national Journal of Pure and Applied Mathematics vol 40 no 1pp 115ndash122 2007

[8] R Kress Linear Integral Equations vol 82 ofAppliedMathemat-ical Sciences Springer Berlin Germany 1989

[9] B L Moiseiwitsch Integral Equations Dover New York NYUSA 2005

[10] A D Polyanin Handbook of Integral Equations CRC PressBoca Raton Fla USA 1998

[11] A Vahidian Kamyad MMehrabinezhad and J Saberi-NadjafildquoA numerical approach for solving linear and nonlinear volterraintegral equations with controlled errorrdquo IAENG InternationalJournal of Applied Mathematics vol 40 no 2 pp 69ndash75 2010

[12] J Goncerzewicz H Marcinkowska W Okrasinski and KTabisz ldquoOn the percolation of water from a cylindrical reservoirinto the surrounding soilrdquo Zastosowania Matematyki vol 16no 2 pp 249ndash261 1978

[13] D Wei ldquoUniqueness of solutions for a class of non-linearvolterra integral equations without continuityrdquo Applied Math-ematics and Mechanics vol 18 no 12 pp 1191ndash1196 1997

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 ISRN Applied Mathematics

119910(119909) = sum119899+1119896=minus1

1198881198961198791ℎ(119909 minus 119909

119896) plugging into the integral equa-

tion we arrive at

1198880= 119892 (119909

0)

119892 (119909119896) = 119888119896minus 120582(120572

21198901198880 + 1205731198901198881) ℎ119870 (119909

0 119909119896)

minus119896minus1

sum119894=1

120582 (120573119890119888119894minus1 + 120572119890119888119894 + 120573119890119888119894+1) ℎ119870 (119909119894 119909119896)

minus 120582 (120573119890119888119896minus1 + 1205722119890119888119896) ℎ119870 (119909

119896 119909119896)

(119896 = 1 2 10)

(28)

The solution is [1198880 1198881 1198882 1198883 1198884 1198885 1198886 1198887 1198888 1198889

11988810] = [009531139 018232333 02623661939 03364740924

04054667723 04700050464 05306294034 0587787554806418545277 06931475935]

Compared with the exact solution [119910(1199090) 119910(119909

1) 119910(119909

2)

119910(1199093) 119910(119909

4) 119910(119909

5) 119910(119909

6) 119910(119909

7) 119910(119909

8) 119910(119909

9) 119910(119909

10)] =

[009531017980 01823215567 02623642644 0336472236604054651081 04700036292 05306282510 0587786664906418538861 06931471805] The error 119864 lt 0000001

Example 6 Consider a system of Volterra integral equations

119906 (119909) = sin119909 minus 119909 + int119909

0

(1199062 (119905) + V2 (119905)) 119889119905

V (119909) = cos119909 minus 12sin2119909 + int

119909

0

(119906 (119905) V (119905)) 119889119905(29)

The exact solutions are 119906(119909) = sin119909 V(119909) = cos119909In an interval (0 1) we let ℎ = 110 119909

119894= 119894ℎ 119894 =

0 1 10 Furthermore let

119906 (119909) =119899

sum119896=0

11988811989611198791ℎ(119909 minus 119909

119896)

V (119909) =119899

sum119896=0

11988811989621198791ℎ(119909 minus 119909

119896)

(119906(119909))2 =119899

sum119896=0

(1198881198961)2 1198791ℎ(119909 minus 119909

119896)

(V(119909))2 =119899

sum119896=0

(1198881198962)2 1198791ℎ(119909 minus 119909

119896)

(119906 (119909) V (119909)) =119899

sum119896=0

(11988811989611198881198962) 1198791ℎ(119909 minus 119909

119896)

119870 (119909 119905) = 1 =119899

sum119894=0

119899

sum119895=0

1198791ℎ(119909 minus 119909

119894) 1198791ℎ(119905 minus 119909

119895)

119892 (119909) = sin119909 minus 119909 =119899

sum119896=0

119892 (119909119896) 1198791ℎ(119909 minus 119909

119896)

ℎ (119909) = cos119909 minus 12sin2119909 =

119899

sum119896=0

ℎ (119909119896) 1198791ℎ(119909 minus 119909

119896)

(30)

plugging into the system we get

11988801

= 0

119892 (119909119896) = 1198881198961

minus (1205722(119888201

+ 119888202) + 120573 (1198882

11+ 119888212)) ℎ

minus119896minus1

sum119894=1

(120573 (1198882119894minus11

+ 1198882119894minus12

) + 120572 (11988821198941+ 11988821198942)

+ 120573 (1198882119894+11

+ 1198882119894+12

)) ℎ

minus (120573 (1198882119896minus11

+ 1198882119896minus12

) + 1205722(11988821198961

+ 11988821198962)) ℎ

(119896 = 1 2 10)

11988802

= 1

ℎ (119909119896) = 1198881198962

minus (1205722(1198880111988802) + 120573 (119888

1111988812)) ℎ

minus119896minus1

sum119894=1

(120573 (119888119894minus11

119888119894minus12

) + 120572 (11988811989411198881198942) + 120573 (119888

119894+11119888119894+12

)) ℎ

minus (120573 (119888119896minus11

119888119896minus12

) + 1205722(11988811989611198881198962)) ℎ

(119896 = 1 2 10) (31)

which is a nearly triangular system of (10 + 1)2 non-linear equations on unknowns [119888

01 11988811 119888

1198991] [11988802 11988812

1198881198992] we need to solve two nonlinear equations 119888

1198941=

1198911(1198881198941 1198881198942) + 119882

1198941 1198881198942

= 1198912(1198881198941 1198881198942) + 119882

1198942 where 119882

11989411198821198941

are numbers each time for 119894 = 0 1 119899Solutions are [119888

01 11988811 119888

1198991] [11988802 11988812 119888

1198992] =

[009983172966 01986591558 02954877407 0389342381604792769003 05643832656 06438004752 071672303870782409006 08405334929] [09949873745 0979999248109551843698 09207890215 08771545950 0824714253307639887345 06955813657 06201723633 05443657395]Compared with the exact solution [0 00998334166401986693307 02955202066 03894183423 0479425538605646424733 06442176872 07173560908 0783326909608414709848] [1 09950041652 0980066577809553364891 09210609940 08775825618 0825335614907648421872 06967067093 06216099682 05403023058]The error lt0005

6 Conclusions

The proposed method is a simple and effective procedurefor solving nonlinear Volterra integral equations as well as a

ISRN Applied Mathematics 7

system of nonlinear Volterra integral equationsThemethodscan be adapted easily to the Volterra integral equations of thefirst kind which have the form 119892(119909) = int

119860

119870(119909 119905)119910(119905)119889119905 Theconvergence rate could be higher if we use more complicatedorthonormal or cardinal splines Nevertheless the resultingsystem of coefficients will be more complicated nonlinearsystems which could take more time and effort to solve

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

The work was funded by the Natural Science Foundationof Anhui Province of China under Grant no 1208085MA15and the Major Project of the Nature Science Foundation ofthe Education Department Anhui Province under Grant noKJ2014ZD30

References

[1] I J Schoenberg ldquoOn trigonometric spline interpolationrdquo Jour-nal of Mathematics and Mechanics vol 13 pp 795ndash825 1964

[2] T Lyche L L Schumaker and S Stanley ldquoQuasi-interpolantsbased on trigonometric splinesrdquo Journal of ApproximationTheory vol 95 no 2 pp 280ndash309 1998

[3] T Lyche and R Winther ldquoA stable recurrence relation fortrigonometric 119861-splinesrdquo Journal of ApproximationTheory vol25 no 3 pp 266ndash279 1979

[4] A Sharma and J Tzimbalario ldquoA class of cardinal trigonometricsplinesrdquo SIAM Journal on Mathematical Analysis vol 7 no 6pp 809ndash819 1976

[5] X Liu ldquoBivariate cardinal spline functions for digital signalprocessingrdquo inTrends in ApproximationTheory K Kopotum TLyche andM Neamtu Eds pp 261ndash271 Vanderbilt UniversityNashville Tenn USA 2001

[6] X Liu ldquoUnivariate and bivariate orthonormal splines and car-dinal splines on compact supportsrdquo Journal of Computationaland Applied Mathematics vol 195 no 1-2 pp 93ndash105 2006

[7] X Liu ldquoInterpolation by cardinal trigonometric splinesrdquo Inter-national Journal of Pure and Applied Mathematics vol 40 no 1pp 115ndash122 2007

[8] R Kress Linear Integral Equations vol 82 ofAppliedMathemat-ical Sciences Springer Berlin Germany 1989

[9] B L Moiseiwitsch Integral Equations Dover New York NYUSA 2005

[10] A D Polyanin Handbook of Integral Equations CRC PressBoca Raton Fla USA 1998

[11] A Vahidian Kamyad MMehrabinezhad and J Saberi-NadjafildquoA numerical approach for solving linear and nonlinear volterraintegral equations with controlled errorrdquo IAENG InternationalJournal of Applied Mathematics vol 40 no 2 pp 69ndash75 2010

[12] J Goncerzewicz H Marcinkowska W Okrasinski and KTabisz ldquoOn the percolation of water from a cylindrical reservoirinto the surrounding soilrdquo Zastosowania Matematyki vol 16no 2 pp 249ndash261 1978

[13] D Wei ldquoUniqueness of solutions for a class of non-linearvolterra integral equations without continuityrdquo Applied Math-ematics and Mechanics vol 18 no 12 pp 1191ndash1196 1997

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

ISRN Applied Mathematics 7

system of nonlinear Volterra integral equationsThemethodscan be adapted easily to the Volterra integral equations of thefirst kind which have the form 119892(119909) = int

119860

119870(119909 119905)119910(119905)119889119905 Theconvergence rate could be higher if we use more complicatedorthonormal or cardinal splines Nevertheless the resultingsystem of coefficients will be more complicated nonlinearsystems which could take more time and effort to solve

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

The work was funded by the Natural Science Foundationof Anhui Province of China under Grant no 1208085MA15and the Major Project of the Nature Science Foundation ofthe Education Department Anhui Province under Grant noKJ2014ZD30

References

[1] I J Schoenberg ldquoOn trigonometric spline interpolationrdquo Jour-nal of Mathematics and Mechanics vol 13 pp 795ndash825 1964

[2] T Lyche L L Schumaker and S Stanley ldquoQuasi-interpolantsbased on trigonometric splinesrdquo Journal of ApproximationTheory vol 95 no 2 pp 280ndash309 1998

[3] T Lyche and R Winther ldquoA stable recurrence relation fortrigonometric 119861-splinesrdquo Journal of ApproximationTheory vol25 no 3 pp 266ndash279 1979

[4] A Sharma and J Tzimbalario ldquoA class of cardinal trigonometricsplinesrdquo SIAM Journal on Mathematical Analysis vol 7 no 6pp 809ndash819 1976

[5] X Liu ldquoBivariate cardinal spline functions for digital signalprocessingrdquo inTrends in ApproximationTheory K Kopotum TLyche andM Neamtu Eds pp 261ndash271 Vanderbilt UniversityNashville Tenn USA 2001

[6] X Liu ldquoUnivariate and bivariate orthonormal splines and car-dinal splines on compact supportsrdquo Journal of Computationaland Applied Mathematics vol 195 no 1-2 pp 93ndash105 2006

[7] X Liu ldquoInterpolation by cardinal trigonometric splinesrdquo Inter-national Journal of Pure and Applied Mathematics vol 40 no 1pp 115ndash122 2007

[8] R Kress Linear Integral Equations vol 82 ofAppliedMathemat-ical Sciences Springer Berlin Germany 1989

[9] B L Moiseiwitsch Integral Equations Dover New York NYUSA 2005

[10] A D Polyanin Handbook of Integral Equations CRC PressBoca Raton Fla USA 1998

[11] A Vahidian Kamyad MMehrabinezhad and J Saberi-NadjafildquoA numerical approach for solving linear and nonlinear volterraintegral equations with controlled errorrdquo IAENG InternationalJournal of Applied Mathematics vol 40 no 2 pp 69ndash75 2010

[12] J Goncerzewicz H Marcinkowska W Okrasinski and KTabisz ldquoOn the percolation of water from a cylindrical reservoirinto the surrounding soilrdquo Zastosowania Matematyki vol 16no 2 pp 249ndash261 1978

[13] D Wei ldquoUniqueness of solutions for a class of non-linearvolterra integral equations without continuityrdquo Applied Math-ematics and Mechanics vol 18 no 12 pp 1191ndash1196 1997

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