Research Article A Reproducing Kernel Hilbert Space Method ...downloads.hindawi.com/journals/aaa/2014/103016.pdf · Research Article A Reproducing Kernel Hilbert Space Method for

Research ArticleA Reproducing Kernel Hilbert Space Method for SolvingSystems of Fractional Integrodifferential Equations

Samia Bushnaq1 Banan Maayah2 Shaher Momani23 and Ahmed Alsaedi3

1 Department of Science King Abdullah II Faculty of Engineering Princess Sumaya University for Technology Amman 11941 Jordan2Department of Mathematics Faculty of Science The University of Jordan Amman 11942 Jordan3Nonlinear Analysis and Applied Mathematics (NAAM) Research Group Faculty of Science King Abdulaziz UniversityJeddah 21589 Saudi Arabia

Correspondence should be addressed to Shaher Momani smomanijuedujo

Received 5 January 2014 Accepted 14 February 2014 Published 24 March 2014

Academic Editor Hossein Jafari

Copyright copy 2014 Samia Bushnaq et alThis is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

We present a new version of the reproducing kernel Hilbert space method (RKHSM) for the solution of systems of fractionalintegrodifferential equations In this approach the solution is obtained as a convergent series with easily computable componentsSeveral illustrative examples are given to demonstrate the effectiveness of the present method The method described in this paperis expected to be further employed to solve similar nonlinear problems in fractional calculus

1 Introduction

In this paper we consider the following system of fractionalintegrodifferential equations

119863120572119894119909119894(119905)

= 119865119894(119905 1199091(119905) 119909

(119896)

1(119905) 119909

119894minus1(119905) 119909

(119896)

119894minus1(119905)

119909119894+1(119905) 119909

(119896)

119894+1(119905) 119909

119899(119905) 119909

(119896)

119899(119905))

+ int

119905

0

119866119894(119905 120591 119909

1(120591) 119909

(119896)

1(120591) 119909

119899(120591)

119909(119896)

119899(120591)) 119889120591

(1)

where 119894 = 1 119899 119896 = 0 1 119898 0 le 119905 le 1 and 119863120572119894 isderivative of order 120572

119894in the sense of Caputo and119898minus1 lt 120572

119894le

119898 subject to the initial conditions

119909(119895)

119894(119886) = 119886

119895119894 119895 = 0 1 119898 minus 1 119894 = 1 2 119899 119886 ge 0

(2)

In the last two decades fractional calculus has founddiverse applications in various scientific and technologi-cal fields [1 2] such as thermal engineering acoustics

electromagnetism control robotics viscoelasticity diffu-sion edge detection turbulence signal processing andmany other physical and biological processes Fractional dif-ferential equations have also been applied in modelingmany physical and engineering problems Most systems offractional integrodifferential equations do not have exactsolutions so numerical techniques are used to solve suchsystems The homotopy perturbation method the Adomiandecomposition method and other methods are used to givean approximate solution to linear and nonlinear problemssee [3ndash13] and the references therein

In our previous work [14] we proposed a reproducingkernel Hilbert space method for solving integrodifferentialequations of fractional order based on the reproducing kerneltheory [14 15] In this paper we will generalize the idea ofthe RKHSM to provide a numerical solution for systems offractional integrodifferential equations (1) To demonstratethe effectiveness of the RKHSM algorithm several numericalexperiments of linear and nonlinear systems of fractionalequations (1) will be presented

This paper is organized as follows An introduction of thealgorithm for solving systems of fractional integrodifferentialequations is given in Section 2 In Section 3 we introduceseveral examples to show the efficiency of themethod Finallya conclusion is given in Section 4

Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014 Article ID 103016 6 pageshttpdxdoiorg1011552014103016

2 Abstract and Applied Analysis

2 The Algorithm

After homogenizing the initial conditions (2) we apply theoperator 119868120572119894 the Riemann-Liouville fractional integral oforder 120572

119894[2 16ndash20] to both sides of (1) to have

119909119894(119905) = 119872

119894(119905) 119894 = 1 2 119899 (3)

where119872119894(119905)

= 119868120572119894 (119865119894(119905 1199091(119905) 119909

(119896)

1(119905) 119909

119894minus1(119905) 119909

(119896)

119894minus1(119905)

119909119894+1(119905) 119909

(119896)

119894+1(119905) 119909

119899(119905) 119909

(119896)

119899(119905))

+ int

119905

0

119866119894(119905 120591 119909

1(120591) 119909

(119896)

1(120591) 119909

119899(120591)

119909(119896)

119899(120591)) 119889120591)

119894 = 1 2 119899 119896 = 0 1 119898

(4)

It is clear that (3) is equivalent to (1) so every solutionof the integral equation (3) is also a solution of our originalproblem (1) and vice versa

To solve (3) by means of the reproducing kernel Hilbertspace method first we need to construct a reproducingkernel of certain spaces119882119898+1

2[119886 119887] = 119906 | 119906

(119895) is absolutelycontinuous 119895 = 1 2 119898 minus 1 and 119906(119898) isin 1198712[119886 119887] in whichevery function satisfies the homogenous initial conditions of(1)

(i) The inner product of the space 11988212[0 1] = 119906 | 119906

is absolutely continuous real value function 1199061015840 isin1198712

[0 1] is given by

⟨119906 V⟩1198821

2

= int

1

0

(119906 (119905) V (119905) + 1199061015840 (119905) V1015840 (119905)) 119889119905 (5)

and norm ||119906||1198821

2

= radic⟨119906 119906⟩1198821

2

In [21] Li and Cui proved that11988212[0 1] is a reproduc-

ing kernel Hilbert space and its reproducing kernel isgiven by

119877 (119909 119910) =1

2 sinh 1

times [cosh (119909 + 119910 minus 1) + cosh 1003816100381610038161003816119909 minus 1199101003816100381610038161003816 minus 1]

(6)

(ii) The inner product of the space11988222[0 1] = 119906 | 119906 119906

1015840

are absolutely continuous real value functions 11990610158401015840 isin1198712

[0 1] 119906(0) = 0 is given by

⟨119906 V⟩1198822

2

= 119906 (0) V (0) + 1199061015840 (0) V1015840 (0)

+ int

1

0

11990610158401015840

(119905) V10158401015840 (119905) 119889119905(7)

and norm ||119906||1198822

2

= radic⟨119906 119906⟩1198822

2

11988222[0 1] is a repro-

ducing kernel Hilbert space and its reproducingkernel is given by

119878 (119909 119910) =

1

6119910 (minus119910

2

+ 3119909 (2 + 119910)) 119910 le 119909

1

6119909 (minus119909

2

+ 3119910 (2 + 119909)) 119910 gt 119909

(8)

(iii) The inner product of the space 11988232[0 1] = 119906 |

119906 1199061015840

11990610158401015840 are absolutely continuous real value func-

tions 119906101584010158401015840 isin 1198712[0 1] 119906(0) = 1199061015840(0) = 0 is given by

⟨119906 V⟩1198823

2

=

2

sum

119894=0

119906(119894)

(0) V(119894) (0) + int1

0

119906101584010158401015840

(119905) V101584010158401015840 (119905) 119889119905 (9)

and norm ||119906||1198823

2

= radic⟨119906 119906⟩1198823

2

11988232[0 1] is a

reproducing kernel Hilbert space and its reproducingkernel is given by

119872(119909 119910) = 119891 (119909 119910) 119910 le 119909

119891 (119910 119909) 119910 gt 119909(10)

where 119891(119909 119910) = (1120)1199102(minus1199092(minus126 + 10119909 minus 51199092 + 1199093) +5(minus1 + 119909)119909119910

2

minus (minus1 + 1199092

)1199103

)The method of obtaining the reproducing kernel can be

found in [15]Let 119871

119894 119882119898+1

2[0 1] rarr 119882

1

2[0 1] such that 119871

119894119909119894(119905) =

119909119894(119905) Then 119871

119894 119894 = 1 2 119899 are bounded linear operators

Let 119905119895infin

119895=1

be a countable dense set in [0 1] Let 120593119894119895(119905) =

119877(119905119895 119905) and 120595119894

119895(119905) = 119871

lowast

119894120593119894

119895(119905) where 119871lowast

119894is the adjoint operator

of 119871119894By Gram-Schmidt process we can construct an orthonor-

mal system 120595119894119895(119905)infin

119895=1

of119882119898+12[0 1] where

120595119894

119895(119905) =

119895

sum

119896=1

120573119894

119895119896120595119894

119896(119905) 120573

119894

119895119895gt 0

forall119895 = 1 2 119894 = 1 2 119899

(11)

Theorem 1 Let 119905119895infin

119895=1be a dense set in [0 1] Then 120595119894

119895(119905)infin

119895=1

is a complete system of119882119898+12[0 1]

For the proof see [14]


119895=1be a dense set in [0 1] and the solution

of (3) is unique on119882119898+12[0 1]Then the solution of (3) is given

by 119909119894(119905) = sum

infin

119895=1119860119895120595119894

119895(119905) where 119860

119895= sum119895

119896=1120573119894

119895119896119872119894(119905119896)

For the proof see [14]One can get an approximate solution 119909

119894119899(119905) by taking

finitely many terms in the series representation of 119909119894(119905) and

119909119894119899(119905) = sum

119899

119895=1119860119895120595119894

119895(119905)

Since 119882119898+1

2[0 1] is a Hilbert space then

suminfin

119895=1suminfin

119896=1120573119894

119895119896119872119894(119905119896) lt infin

Abstract and Applied Analysis 3

02 04 06 08 10

15

20

25

30

35

40x(t)

t

120572 = 1

120572 = 09

120572 = 08

120572 = 07

(a)

02 04 06 08 10

000005

000010

000015

000020

000025

000030

Abso

lute

erro

r

t

(b)

02 04 06 08 10t

y(t)

minus05

minus10

minus15

120572 = 1

120572 = 09

120572 = 08

120572 = 07

(c)

Abso

lute

erro

r

02 04 06 08 10

000005

000010

000015

000020

000025

000030

t

(d)

Figure 1 Graphical results for Example 1 when 1205721= 1205722= 120572 = 1 09 08 and 07

Theorem 3 The approximate solution 119909119894119899(119905) and its deriva-

tives 119909(119895)119894119899

are uniformly convergent to 119909(119895)119894(119905) 119894 = 1 2 119899

119895 = 0 1

Proof By the reproducing kernel property of 119870(119909 119910) andSchwarz inequality we can obtain

1003816100381610038161003816119909119894119899 (119905) minus 119909119894 (119905)1003816100381610038161003816 =100381610038161003816100381610038161003816⟨119909119894119899(119905) minus 119909

119894(119905) 119870 (119909 119910)⟩

119882119898+1

2

100381610038161003816100381610038161003816

le1003817100381710038171003817119870 (119909 119910)

1003817100381710038171003817119882119898+12

1003817100381710038171003817119909119894119899 (119905) minus 119909119894 (119905)1003817100381710038171003817119882119898+12

le 1198880

1003817100381710038171003817119909119894119899 (119905) minus 119909119894 (119905)1003817100381710038171003817119882119898+12

(12)

where 1198880is a constant

By the representation of119870(119909 119910) we can obtain

100381610038161003816100381610038161003816119909(119895)

119894119899(119905) minus 119909

(119895)

119894(119905)100381610038161003816100381610038161003816

=

1003816100381610038161003816100381610038161003816⟨119909(119895)

119894119899(119905) minus 119909

(119895)

119894(119905) 119870

(119895)

(119909 119910)⟩119882119898+1

2

1003816100381610038161003816100381610038161003816

le10038171003817100381710038171003817119870(119895)

(119909 119910)10038171003817100381710038171003817119882119898+12

100381710038171003817100381710038171003817119909(119895)

119894119899(119905) minus 119909

(119895)

119894(119905)100381710038171003817100381710038171003817119882119898+12

(13)

Since 119870(119895)(119909 119910) 119895 = 1 2 is uniformly bounded about 119909and 119910 we have

10038171003817100381710038171003817119870(119895)

(119909 119910)10038171003817100381710038171003817119882119898+12

le 119888119895 119895 = 1 2 (14)


02 04 06 08 10

02

04

06

08

10

12

x(t)

t

120572 = 1

120572 = 09

120572 = 08

120572 = 07

(a)

02 04 06 08 10

00005

00010

00015

Abso

lute

erro

r

t

(b)

02 04 06 08 10

11

12

13

14

15

16

17

y(t)

120572 = 1

120572 = 09

120572 = 08

120572 = 07

t

(c)

Abso

lute

erro

r

02 04 06 08 10

000005

000010

000015

t

(d)


and so

100381610038161003816100381610038161003816119909(119895)

119894119899(119905) minus 119909

(119895)

119894(119905)100381610038161003816100381610038161003816le 119888119895

100381710038171003817100381710038171003817119909(119895)

119894119899(119905) minus 119909

(119895)

119894(119905)100381710038171003817100381710038171003817119882119898+12

119895 = 1 2

(15)

Thus 119909119894(119905) and its derivatives 119909(119895)

119894119899(119905) are uniformly convergent

to 119909(119895)119894(119905) 119895 = 1 2

3 Numerical Results

In this paper three numerical examples are given to show theaccuracy of this methodThe computations are performed byMathematica 80We compare the results by thismethodwiththe exact solution of each example

Example 1 Consider the following linear system of fractionalintegrodifferential equations

1198631205721119909 (119905) = 1 + 119905 + 119905

2

minus 119910 (119905) minus int

119905

0

(119909 (120591) + 119910 (120591)) 119889120591

1198631205722119909 (119905) = minus1 minus 119905 + 119909 (120591) minus int

119905

0

(119909 (120591) minus 119910 (120591)) 119889120591

119909 (0) = 1 119910 (0) = minus1 0 lt 1205721 1205722le 1

(16)

The exact solution for1205721= 1205722= 1 is119909(119905) = 119905+119890119905 119910(119905) = 119905minus 119890119905

After homogenizing the initial conditions and using thismethod taking 119905

119894= 119894119899 119894 = 1 2 119899 and 119899 = 20 the graphs

of the approximate solutions for different values of 1205721and

1205722are plotted in Figure 1 From Figure 1 it is clear that the

approximate solutions are in good agreement with the exactsolutions when 120572

1= 1205722= 1 and the solution continuously

depends on the fractional derivative


02 04 06 08 10

15

20

25

30

35

x(t)

t

120572 = 2

120572 = 19

120572 = 18

120572 = 17

(a)

02 04 06 08 10

Abso

lute

erro

r

t

8 times 10minus6

6 times 10minus6

4 times 10minus6

2 times 10minus6

(b)

02 04 06 08 10

minus10

minus12

minus14

minus16

minus18

t

y(t)

120572 = 2

120572 = 19

120572 = 18

120572 = 17

(c)

02 04 06 08 10

Abso

lute

erro

r

t

1 times 10minus6

8 times 10minus7

6 times 10minus7

4 times 10minus7

2 times 10minus7

(d)


Example 2 Consider the following nonlinear system offractional integrodifferential equations

1198631205721119909 (119905) = 1 minus

1

211991010158402

(119905) + int

119905

0

[(119905 minus 120591) 119910 (120591) + 119909 (120591) 119910 (120591)] 119889120591

1198631205722119909 (119905) = 2119905 + int

119905

0

[(119905 minus 120591) 119909 (120591) minus 1199102

(120591) + 1199092

(120591)] 119889120591

119909 (0) = 0 119910 (0) = 1 0 lt 1205721 1205722le 1

(17)

The exact solution for 1205721= 1205722= 1 is 119909(119905) = sinh 119905 119910(119905) =

cosh 119905After homogenizing the initial conditions and using this

method taking 119905119894= 119894119899 119894 = 1 2 119899 and 119899 = 30 the graphs

of the approximate solutions for different values of 1205721and 120572

2

are plotted in Figure 2


1198631205721119909 (119905) = 1 minus

1199053

3minus1

211991010158402

(119905) +1

2int

119905

0

(1199092

(120591) + 1199102

(120591)) 119889120591

1198631205722119909 (119905) = minus1 + 119905

2

minus 119905119909 (119905) +1

4int

119905

0

(1199092

(120591) minus 1199102

(120591)) 119889120591

119909 (0) = 1 1199091015840

(0) = 2 119910 (0) = minus1

1199101015840

(0) = 0 1 lt 1205721 1205722le 2

(18)

The exact solution for 1205721= 1205722= 2 is 119909(119905) = 119905 + 1198903 119910(119905) =

119905 minus 119890119905After homogenizing the initial conditions and using this




2


4 Conclusion

In this paper we introduce a new algorithm for solving sys-tems of fractional integrodifferential equations The approxi-mate solution obtained by this method and its derivative areboth uniformly convergentThe obtained results demonstratethe reliability of the algorithm and its wider applicabilityto linear and nonlinear systems of fractional differentialequations

Conflict of Interests

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

References

[1] I Podlubuy Fractional Differential Equations Academic PressNew York NY USA 1999

[2] F Mainardi ldquoFractional calculus some basic problems incontinuum and statistical mechanics rdquo in Fractals and Frac-tional Calculus in Continuum Mechanics A Carpinteri and FMainardi Eds Springer New York NY USA 1997

[3] V S Erturk and S Momani ldquoSolving systems of fractionaldifferential equations using differential transform methodrdquoJournal of Computational and AppliedMathematics vol 215 no1 pp 142ndash151 2008

[4] J Duan J An and M Xu ldquoSolution of system of fractionaldifferential equations by Adomian decomposition methodrdquoApplied Mathematics vol 22 no 1 pp 7ndash12 2007

[5] A S Mohamed and R A Mahmoud ldquoAn algorithm for thenumerical solution of systems of fractional differential equa-tionsrdquo International Journal of Computer Applications vol 65no 11 2013

[6] S Momani and R Qaralleh ldquoAn efficient method for solvingsystems of fractional integro-differential equationsrdquo Computersamp Mathematics with Applications vol 52 no 3-4 pp 459ndash4702006

[7] M Zurigat S Momani and A Alawneh ldquoHomotopy analysismethod for systems of fractional integro-differential equationsrdquoin Proceeding of the 4th International Workshop of AdvancedComputation for Engineering Applications pp 106ndash111 2008

[8] M Zurigat SMomani Z Odibat andA Alawneh ldquoThe homo-topy analysis method for handling systems of fractional differ-ential equationsrdquo Applied Mathematical Modelling vol 34 no1 pp 24ndash35 2010

[9] S Momani and Z Odibat ldquoNumerical approach to differentialequations of fractional orderrdquo Journal of Computational andApplied Mathematics vol 207 no 1 pp 96ndash110 2007

[10] O Abdulaziz I Hashim and S Momani ldquoSolving systems offractional differential equations by homotopy-perturbationmethodrdquo Physics Letters A General Atomic and Solid StatePhysics vol 372 no 4 pp 451ndash459 2008

[11] Y Lin and Y Zhou ldquoSolving nonlinear pseudoparabolic equa-tions with nonlocal boundary conditions in reproducing kernelspacerdquo Numerical Algorithms vol 52 no 2 pp 173ndash186 2009

[12] B Ghazanfari and A G Ghazanfari ldquoSolving system of frac-tional differential equations by fractional complex transformmethodrdquoAsian Journal of Applied Sciences vol 5 no 6 pp 438ndash444 2012

[13] R K Saeed and H M Sadeq ldquoSolving a system of linear fred-holm fractional integro-differential equations using homotopyperturbation methodrdquo Australian Journal of Basic and AppliedSciences vol 4 no 4 pp 633ndash638 2010

[14] S Bushnaq S Momani and Y Zhon ldquoA reproducing KernelHilbert space method for solving integro-differential equationsof fractional orderrdquo Journal of Optimization Theory and Appli-cations vol 156 no 1 pp 96ndash105 2012

[15] M Cui and Y Lin Nonlinear Numerical Analysis in the Repro-ducing Kernel Hilbert Space Nova Science Publishers NewYork NY USA 2009

[16] J S Leszczynski An Introduction to Fractional MechanicsCzestochowa University of Technology Czestochowa Poland2011

[17] M Caputo ldquoLinear Models of Dissipation whose Q is almostFrequency Independent-IIrdquo Geophysical Journal of the RoyalAstronomical Society vol 13 no 5 pp 529ndash539 1967

[18] A A Kilbas H M Srivastava and J J Trujillo Theory andApplications of Fractional Differential Equations Elsevier Ams-terdam The Netherlands 2006

[19] R Hilfer Applications of Fractional Calculus in Physics WorldScientific River Edge NJ USA 2000

[20] S G Samko A A Kilbas andO IMarichev Fractional Integraland Derivatives Theory and Applications Gordon and BreachScience Publishers Yverdon Switzerland 1993

[21] C-l Li and M-G Cui ldquoThe exact solution for solving a classnonlinear operator equations in the reproducing kernel spacerdquoApplied Mathematics and Computation vol 143 no 2-3 pp393ndash399 2003

Submit your manuscripts athttpwwwhindawicom

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Stochastic AnalysisInternational Journal of


2 The Algorithm

After homogenizing the initial conditions (2) we apply theoperator 119868120572119894 the Riemann-Liouville fractional integral oforder 120572

119894[2 16ndash20] to both sides of (1) to have

119909119894(119905) = 119872

119894(119905) 119894 = 1 2 119899 (3)

where119872119894(119905)

= 119868120572119894 (119865119894(119905 1199091(119905) 119909

(119896)

1(119905) 119909

119894minus1(119905) 119909

(119896)

119894minus1(119905)

119909119894+1(119905) 119909

(119896)

119894+1(119905) 119909

119899(119905) 119909

(119896)

119899(119905))

+ int

119905

0

119866119894(119905 120591 119909

1(120591) 119909

(119896)

1(120591) 119909

119899(120591)

119909(119896)

119899(120591)) 119889120591)

119894 = 1 2 119899 119896 = 0 1 119898

(4)

It is clear that (3) is equivalent to (1) so every solutionof the integral equation (3) is also a solution of our originalproblem (1) and vice versa

To solve (3) by means of the reproducing kernel Hilbertspace method first we need to construct a reproducingkernel of certain spaces119882119898+1

2[119886 119887] = 119906 | 119906

(119895) is absolutelycontinuous 119895 = 1 2 119898 minus 1 and 119906(119898) isin 1198712[119886 119887] in whichevery function satisfies the homogenous initial conditions of(1)

(i) The inner product of the space 11988212[0 1] = 119906 | 119906

is absolutely continuous real value function 1199061015840 isin1198712

[0 1] is given by

⟨119906 V⟩1198821

2

= int

1

0

(119906 (119905) V (119905) + 1199061015840 (119905) V1015840 (119905)) 119889119905 (5)

and norm ||119906||1198821

2

= radic⟨119906 119906⟩1198821

2

In [21] Li and Cui proved that11988212[0 1] is a reproduc-

ing kernel Hilbert space and its reproducing kernel isgiven by

119877 (119909 119910) =1

2 sinh 1

times [cosh (119909 + 119910 minus 1) + cosh 1003816100381610038161003816119909 minus 1199101003816100381610038161003816 minus 1]

(6)

(ii) The inner product of the space11988222[0 1] = 119906 | 119906 119906

1015840

are absolutely continuous real value functions 11990610158401015840 isin1198712

[0 1] 119906(0) = 0 is given by

⟨119906 V⟩1198822

2

= 119906 (0) V (0) + 1199061015840 (0) V1015840 (0)

+ int

1

0

11990610158401015840

(119905) V10158401015840 (119905) 119889119905(7)

and norm ||119906||1198822

2

= radic⟨119906 119906⟩1198822

2

11988222[0 1] is a repro-

ducing kernel Hilbert space and its reproducingkernel is given by

119878 (119909 119910) =

1

6119910 (minus119910

2

+ 3119909 (2 + 119910)) 119910 le 119909

1

6119909 (minus119909

2

+ 3119910 (2 + 119909)) 119910 gt 119909

(8)

(iii) The inner product of the space 11988232[0 1] = 119906 |

119906 1199061015840

11990610158401015840 are absolutely continuous real value func-

tions 119906101584010158401015840 isin 1198712[0 1] 119906(0) = 1199061015840(0) = 0 is given by

⟨119906 V⟩1198823

2

=

2

sum

119894=0

119906(119894)

(0) V(119894) (0) + int1

0

119906101584010158401015840

(119905) V101584010158401015840 (119905) 119889119905 (9)

and norm ||119906||1198823

2

= radic⟨119906 119906⟩1198823

2

11988232[0 1] is a

reproducing kernel Hilbert space and its reproducingkernel is given by

119872(119909 119910) = 119891 (119909 119910) 119910 le 119909

119891 (119910 119909) 119910 gt 119909(10)

where 119891(119909 119910) = (1120)1199102(minus1199092(minus126 + 10119909 minus 51199092 + 1199093) +5(minus1 + 119909)119909119910

2

minus (minus1 + 1199092

)1199103

)The method of obtaining the reproducing kernel can be

found in [15]Let 119871

119894 119882119898+1

2[0 1] rarr 119882

1

2[0 1] such that 119871

119894119909119894(119905) =

119909119894(119905) Then 119871

119894 119894 = 1 2 119899 are bounded linear operators

Let 119905119895infin

119895=1

be a countable dense set in [0 1] Let 120593119894119895(119905) =

119877(119905119895 119905) and 120595119894

119895(119905) = 119871

lowast

119894120593119894

119895(119905) where 119871lowast

119894is the adjoint operator

of 119871119894By Gram-Schmidt process we can construct an orthonor-

mal system 120595119894119895(119905)infin

119895=1

of119882119898+12[0 1] where

120595119894

119895(119905) =

119895

sum

119896=1

120573119894

119895119896120595119894

119896(119905) 120573

119894

119895119895gt 0

forall119895 = 1 2 119894 = 1 2 119899

(11)


119895=1be a dense set in [0 1] Then 120595119894

119895(119905)infin

119895=1

is a complete system of119882119898+12[0 1]

For the proof see [14]


119895=1be a dense set in [0 1] and the solution

of (3) is unique on119882119898+12[0 1]Then the solution of (3) is given

by 119909119894(119905) = sum

infin

119895=1119860119895120595119894

119895(119905) where 119860

119895= sum119895

119896=1120573119894

119895119896119872119894(119905119896)

For the proof see [14]One can get an approximate solution 119909

119894119899(119905) by taking

finitely many terms in the series representation of 119909119894(119905) and

119909119894119899(119905) = sum

119899

119895=1119860119895120595119894

119895(119905)

Since 119882119898+1

2[0 1] is a Hilbert space then

suminfin

119895=1suminfin

119896=1120573119894

119895119896119872119894(119905119896) lt infin


02 04 06 08 10

15

20

25

30

35

40x(t)

t

120572 = 1

120572 = 09

120572 = 08

120572 = 07

(a)

02 04 06 08 10

000005

000010

000015

000020

000025

000030

Abso

lute

erro

r

t

(b)

02 04 06 08 10t

y(t)

minus05

minus10

minus15

120572 = 1

120572 = 09

120572 = 08

120572 = 07

(c)

Abso

lute

erro

r

02 04 06 08 10

000005

000010

000015

000020

000025

000030

t

(d)



tives 119909(119895)119894119899


119895 = 0 1


1003816100381610038161003816119909119894119899 (119905) minus 119909119894 (119905)1003816100381610038161003816 =100381610038161003816100381610038161003816⟨119909119894119899(119905) minus 119909

119894(119905) 119870 (119909 119910)⟩

119882119898+1

2

100381610038161003816100381610038161003816

le1003817100381710038171003817119870 (119909 119910)

1003817100381710038171003817119882119898+12

1003817100381710038171003817119909119894119899 (119905) minus 119909119894 (119905)1003817100381710038171003817119882119898+12

le 1198880

1003817100381710038171003817119909119894119899 (119905) minus 119909119894 (119905)1003817100381710038171003817119882119898+12

(12)



100381610038161003816100381610038161003816119909(119895)

119894119899(119905) minus 119909

(119895)

119894(119905)100381610038161003816100381610038161003816

=

1003816100381610038161003816100381610038161003816⟨119909(119895)

119894119899(119905) minus 119909

(119895)

119894(119905) 119870

(119895)

(119909 119910)⟩119882119898+1

2

1003816100381610038161003816100381610038161003816

le10038171003817100381710038171003817119870(119895)

(119909 119910)10038171003817100381710038171003817119882119898+12

100381710038171003817100381710038171003817119909(119895)

119894119899(119905) minus 119909

(119895)

119894(119905)100381710038171003817100381710038171003817119882119898+12

(13)


10038171003817100381710038171003817119870(119895)

(119909 119910)10038171003817100381710038171003817119882119898+12

le 119888119895 119895 = 1 2 (14)


02 04 06 08 10

02

04

06

08

10

12

x(t)

t

120572 = 1

120572 = 09

120572 = 08

120572 = 07

(a)

02 04 06 08 10

00005

00010

00015

Abso

lute

erro

r

t

(b)

02 04 06 08 10

11

12

13

14

15

16

17

y(t)

120572 = 1

120572 = 09

120572 = 08

120572 = 07

t

(c)

Abso

lute

erro

r

02 04 06 08 10

000005

000010

000015

t

(d)


and so

100381610038161003816100381610038161003816119909(119895)

119894119899(119905) minus 119909

(119895)

119894(119905)100381610038161003816100381610038161003816le 119888119895

100381710038171003817100381710038171003817119909(119895)

119894119899(119905) minus 119909

(119895)

119894(119905)100381710038171003817100381710038171003817119882119898+12

119895 = 1 2

(15)



to 119909(119895)119894(119905) 119895 = 1 2

3 Numerical Results



1198631205721119909 (119905) = 1 + 119905 + 119905

2


119905

0

(119909 (120591) + 119910 (120591)) 119889120591


119905

0

(119909 (120591) minus 119910 (120591)) 119889120591

119909 (0) = 1 119910 (0) = minus1 0 lt 1205721 1205722le 1

(16)



119894= 119894119899 119894 = 1 2 119899 and 119899 = 20 the graphs







02 04 06 08 10

15

20

25

30

35

x(t)

t

120572 = 2

120572 = 19

120572 = 18

120572 = 17

(a)

02 04 06 08 10

Abso

lute

erro

r

t

8 times 10minus6

6 times 10minus6

4 times 10minus6

2 times 10minus6

(b)

02 04 06 08 10

minus10

minus12

minus14

minus16

minus18

t

y(t)

120572 = 2

120572 = 19

120572 = 18

120572 = 17

(c)

02 04 06 08 10

Abso

lute

erro

r

t

1 times 10minus6

8 times 10minus7

6 times 10minus7

4 times 10minus7

2 times 10minus7

(d)



1198631205721119909 (119905) = 1 minus

1

211991010158402

(119905) + int

119905

0

[(119905 minus 120591) 119910 (120591) + 119909 (120591) 119910 (120591)] 119889120591

1198631205722119909 (119905) = 2119905 + int

119905

0

[(119905 minus 120591) 119909 (120591) minus 1199102

(120591) + 1199092

(120591)] 119889120591

119909 (0) = 0 119910 (0) = 1 0 lt 1205721 1205722le 1

(17)





2



1198631205721119909 (119905) = 1 minus

1199053

3minus1

211991010158402

(119905) +1

2int

119905

0

(1199092

(120591) + 1199102

(120591)) 119889120591

1198631205722119909 (119905) = minus1 + 119905

2

minus 119905119909 (119905) +1

4int

119905

0

(1199092

(120591) minus 1199102

(120591)) 119889120591

119909 (0) = 1 1199091015840

(0) = 2 119910 (0) = minus1

1199101015840

(0) = 0 1 lt 1205721 1205722le 2

(18)






2


4 Conclusion




References





























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










02 04 06 08 10

15

20

25

30

35

40x(t)

t

120572 = 1

120572 = 09

120572 = 08

120572 = 07

(a)

02 04 06 08 10

000005

000010

000015

000020

000025

000030

Abso

lute

erro

r

t

(b)

02 04 06 08 10t

y(t)

minus05

minus10

minus15

120572 = 1

120572 = 09

120572 = 08

120572 = 07

(c)

Abso

lute

erro

r

02 04 06 08 10

000005

000010

000015

000020

000025

000030

t

(d)



tives 119909(119895)119894119899


119895 = 0 1


1003816100381610038161003816119909119894119899 (119905) minus 119909119894 (119905)1003816100381610038161003816 =100381610038161003816100381610038161003816⟨119909119894119899(119905) minus 119909

119894(119905) 119870 (119909 119910)⟩

119882119898+1

2

100381610038161003816100381610038161003816

le1003817100381710038171003817119870 (119909 119910)

1003817100381710038171003817119882119898+12

1003817100381710038171003817119909119894119899 (119905) minus 119909119894 (119905)1003817100381710038171003817119882119898+12

le 1198880

1003817100381710038171003817119909119894119899 (119905) minus 119909119894 (119905)1003817100381710038171003817119882119898+12

(12)



100381610038161003816100381610038161003816119909(119895)

119894119899(119905) minus 119909

(119895)

119894(119905)100381610038161003816100381610038161003816

=

1003816100381610038161003816100381610038161003816⟨119909(119895)

119894119899(119905) minus 119909

(119895)

119894(119905) 119870

(119895)

(119909 119910)⟩119882119898+1

2

1003816100381610038161003816100381610038161003816

le10038171003817100381710038171003817119870(119895)

(119909 119910)10038171003817100381710038171003817119882119898+12

100381710038171003817100381710038171003817119909(119895)

119894119899(119905) minus 119909

(119895)

119894(119905)100381710038171003817100381710038171003817119882119898+12

(13)


10038171003817100381710038171003817119870(119895)

(119909 119910)10038171003817100381710038171003817119882119898+12

le 119888119895 119895 = 1 2 (14)


02 04 06 08 10

02

04

06

08

10

12

x(t)

t

120572 = 1

120572 = 09

120572 = 08

120572 = 07

(a)

02 04 06 08 10

00005

00010

00015

Abso

lute

erro

r

t

(b)

02 04 06 08 10

11

12

13

14

15

16

17

y(t)

120572 = 1

120572 = 09

120572 = 08

120572 = 07

t

(c)

Abso

lute

erro

r

02 04 06 08 10

000005

000010

000015

t

(d)


and so

100381610038161003816100381610038161003816119909(119895)

119894119899(119905) minus 119909

(119895)

119894(119905)100381610038161003816100381610038161003816le 119888119895

100381710038171003817100381710038171003817119909(119895)

119894119899(119905) minus 119909

(119895)

119894(119905)100381710038171003817100381710038171003817119882119898+12

119895 = 1 2

(15)



to 119909(119895)119894(119905) 119895 = 1 2

3 Numerical Results



1198631205721119909 (119905) = 1 + 119905 + 119905

2


119905

0

(119909 (120591) + 119910 (120591)) 119889120591


119905

0

(119909 (120591) minus 119910 (120591)) 119889120591

119909 (0) = 1 119910 (0) = minus1 0 lt 1205721 1205722le 1

(16)



119894= 119894119899 119894 = 1 2 119899 and 119899 = 20 the graphs







02 04 06 08 10

15

20

25

30

35

x(t)

t

120572 = 2

120572 = 19

120572 = 18

120572 = 17

(a)

02 04 06 08 10

Abso

lute

erro

r

t

8 times 10minus6

6 times 10minus6

4 times 10minus6

2 times 10minus6

(b)

02 04 06 08 10

minus10

minus12

minus14

minus16

minus18

t

y(t)

120572 = 2

120572 = 19

120572 = 18

120572 = 17

(c)

02 04 06 08 10

Abso

lute

erro

r

t

1 times 10minus6

8 times 10minus7

6 times 10minus7

4 times 10minus7

2 times 10minus7

(d)



1198631205721119909 (119905) = 1 minus

1

211991010158402

(119905) + int

119905

0

[(119905 minus 120591) 119910 (120591) + 119909 (120591) 119910 (120591)] 119889120591

1198631205722119909 (119905) = 2119905 + int

119905

0

[(119905 minus 120591) 119909 (120591) minus 1199102

(120591) + 1199092

(120591)] 119889120591

119909 (0) = 0 119910 (0) = 1 0 lt 1205721 1205722le 1

(17)





2



1198631205721119909 (119905) = 1 minus

1199053

3minus1

211991010158402

(119905) +1

2int

119905

0

(1199092

(120591) + 1199102

(120591)) 119889120591

1198631205722119909 (119905) = minus1 + 119905

2

minus 119905119909 (119905) +1

4int

119905

0

(1199092

(120591) minus 1199102

(120591)) 119889120591

119909 (0) = 1 1199091015840

(0) = 2 119910 (0) = minus1

1199101015840

(0) = 0 1 lt 1205721 1205722le 2

(18)






2


4 Conclusion




References





























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










02 04 06 08 10

02

04

06

08

10

12

x(t)

t

120572 = 1

120572 = 09

120572 = 08

120572 = 07

(a)

02 04 06 08 10

00005

00010

00015

Abso

lute

erro

r

t

(b)

02 04 06 08 10

11

12

13

14

15

16

17

y(t)

120572 = 1

120572 = 09

120572 = 08

120572 = 07

t

(c)

Abso

lute

erro

r

02 04 06 08 10

000005

000010

000015

t

(d)


and so

100381610038161003816100381610038161003816119909(119895)

119894119899(119905) minus 119909

(119895)

119894(119905)100381610038161003816100381610038161003816le 119888119895

100381710038171003817100381710038171003817119909(119895)

119894119899(119905) minus 119909

(119895)

119894(119905)100381710038171003817100381710038171003817119882119898+12

119895 = 1 2

(15)



to 119909(119895)119894(119905) 119895 = 1 2

3 Numerical Results



1198631205721119909 (119905) = 1 + 119905 + 119905

2


119905

0

(119909 (120591) + 119910 (120591)) 119889120591


119905

0

(119909 (120591) minus 119910 (120591)) 119889120591

119909 (0) = 1 119910 (0) = minus1 0 lt 1205721 1205722le 1

(16)



119894= 119894119899 119894 = 1 2 119899 and 119899 = 20 the graphs







02 04 06 08 10

15

20

25

30

35

x(t)

t

120572 = 2

120572 = 19

120572 = 18

120572 = 17

(a)

02 04 06 08 10

Abso

lute

erro

r

t

8 times 10minus6

6 times 10minus6

4 times 10minus6

2 times 10minus6

(b)

02 04 06 08 10

minus10

minus12

minus14

minus16

minus18

t

y(t)

120572 = 2

120572 = 19

120572 = 18

120572 = 17

(c)

02 04 06 08 10

Abso

lute

erro

r

t

1 times 10minus6

8 times 10minus7

6 times 10minus7

4 times 10minus7

2 times 10minus7

(d)



1198631205721119909 (119905) = 1 minus

1

211991010158402

(119905) + int

119905

0

[(119905 minus 120591) 119910 (120591) + 119909 (120591) 119910 (120591)] 119889120591

1198631205722119909 (119905) = 2119905 + int

119905

0

[(119905 minus 120591) 119909 (120591) minus 1199102

(120591) + 1199092

(120591)] 119889120591

119909 (0) = 0 119910 (0) = 1 0 lt 1205721 1205722le 1

(17)





2



1198631205721119909 (119905) = 1 minus

1199053

3minus1

211991010158402

(119905) +1

2int

119905

0

(1199092

(120591) + 1199102

(120591)) 119889120591

1198631205722119909 (119905) = minus1 + 119905

2

minus 119905119909 (119905) +1

4int

119905

0

(1199092

(120591) minus 1199102

(120591)) 119889120591

119909 (0) = 1 1199091015840

(0) = 2 119910 (0) = minus1

1199101015840

(0) = 0 1 lt 1205721 1205722le 2

(18)






2


4 Conclusion




References





























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










02 04 06 08 10

15

20

25

30

35

x(t)

t

120572 = 2

120572 = 19

120572 = 18

120572 = 17

(a)

02 04 06 08 10

Abso

lute

erro

r

t

8 times 10minus6

6 times 10minus6

4 times 10minus6

2 times 10minus6

(b)

02 04 06 08 10

minus10

minus12

minus14

minus16

minus18

t

y(t)

120572 = 2

120572 = 19

120572 = 18

120572 = 17

(c)

02 04 06 08 10

Abso

lute

erro

r

t

1 times 10minus6

8 times 10minus7

6 times 10minus7

4 times 10minus7

2 times 10minus7

(d)



1198631205721119909 (119905) = 1 minus

1

211991010158402

(119905) + int

119905

0

[(119905 minus 120591) 119910 (120591) + 119909 (120591) 119910 (120591)] 119889120591

1198631205722119909 (119905) = 2119905 + int

119905

0

[(119905 minus 120591) 119909 (120591) minus 1199102

(120591) + 1199092

(120591)] 119889120591

119909 (0) = 0 119910 (0) = 1 0 lt 1205721 1205722le 1

(17)





2



1198631205721119909 (119905) = 1 minus

1199053

3minus1

211991010158402

(119905) +1

2int

119905

0

(1199092

(120591) + 1199102

(120591)) 119889120591

1198631205722119909 (119905) = minus1 + 119905

2

minus 119905119909 (119905) +1

4int

119905

0

(1199092

(120591) minus 1199102

(120591)) 119889120591

119909 (0) = 1 1199091015840

(0) = 2 119910 (0) = minus1

1199101015840

(0) = 0 1 lt 1205721 1205722le 2

(18)






2


4 Conclusion




References





























Volume 2014




Journal of











Journal of


Function Spaces






Algebra











2


4 Conclusion




References





























Volume 2014




Journal of











Journal of


Function Spaces






Algebra
















Volume 2014




Journal of











Journal of


Function Spaces






Algebra









Documents

Research Article A Reproducing Kernel Hilbert Space Method ...downloads.hindawi.com/journals/aaa/2014/103016.pdf · Research Article A Reproducing Kernel Hilbert Space Method for