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Research ArticleLegendre Polynomials Operational MatrixMethod for Solving Fractional Partial DifferentialEquations with Variable Coefficients
Yongqiang Yang Yunpeng Ma and Lifeng Wang
School of Aeronautic Science and Technology Beihang University Beijing 100191 China
Correspondence should be addressed to Yunpeng Ma mayunpeng088163com
Received 27 January 2015 Accepted 4 May 2015
Academic Editor Francesco Pesavento
Copyright copy 2015 Yongqiang Yang et alThis is an open access article distributed under theCreative CommonsAttribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited
A numerical method for solving a class of fractional partial differential equations with variable coefficients based on Legendrepolynomials is proposed A fractional order operational matrix of Legendre polynomials is also derived The initial equations aretransformed into the products of several matrixes by using the operational matrix A system of linear equations is obtained bydispersing the coefficients and the products of matrixes Only a small number of Legendre polynomials are needed to acquirea satisfactory result Results obtained using the scheme presented here show that the numerical method is very effective andconvenient for solving fractional partial differential equations with variable coefficients
1 Introduction
Thesubject of factional calculuswas found over 300 years agoThe theory of integrals and derivatives of noninteger ordergoes back to Leibnitz Liouville and Letnikov In recent yearsfractional derivative and fractional differential equationshave played a very significant role in many areas in fluid flowphysics mechanics and other applications A lot of practicalproblems can be elegantly modeled with the help of thefractional derivative [1ndash5] Fractional derivatives provide anexcellent instrument for the description ofmemory and here-ditary properties of various materials and processes Due tothe increasing applications a lot of attention has been paid tonumerical and exact solution of fractional differential equa-tions and fractional partial equationsThe analytical solutionsof fractional differential equations are still in a preliminarystage Except in a limited number of these equations we havedifficulty in seeking their analytical as well as numerical solu-tions Thus there have been attempts to develop the methodsfor getting analytical and numerical solutions of fractionaldifferential equations Recently some methods have drawnattention such as Adomian decomposition method (ADM)[6 7] variational iteration method (VIM) [8] generalized
differential transform method (GDTM) [9ndash11] finite differ-ence method (FDM) [12] and wavelet method [13 14]
In this paper our study focuses on a class of fractionalpartial differential equations as follows
119886 (119909)
120597120572
119906 (119909 119905)
120597119909120572
+ 119887 (119909)
120597120573
119906 (119909 119905)
120597119905120573
= 119891 (119909 119905) (1)
such that the initial conditions119906 (119909 0) = 119892 (119909)
119906 (0 119905) = ℎ (119905)
(2)
where 120597120572
119906(119909 119905)120597119909120572 and 120597
120573
119906(119909 119905)120597119905120573 are fractional deriva-
tives of Caputo sense 0 lt 120572 120573 le 1 119891(119909 119905) 119886(119909) and 119887(119909) arethe known and 119906(119909 119905) is the unknown
There have been several methods for solving the frac-tional partial differential equation Doha et al used Jacobi tauapproximation to solve the numerical solution of the spacefractional diffusion equation [15] Yi et al [16] applied blockpulse functions method to obtain the fractional partial equa-tions Podlubny [17] obtained the numerical solution of thefractional partial differential equations with constant coeffi-cients by using Laplace transform method
Hindawi Publishing CorporationMathematical Problems in EngineeringVolume 2015 Article ID 915195 9 pageshttpdxdoiorg1011552015915195
2 Mathematical Problems in Engineering
2 Definitions of FractionalDerivatives and Integrals
Definition 1 Riemann-Liouville fractional integral of order120572 (120572 ge 0) is defined as follows [17]
119868120572
119906 (119905) =
1Γ (120572)
int
119905
0(119905 minus 120591)
120572minus1119906 (120591) 119889120591 119905 gt 0
1198680119906 (119905) = 119906 (119905)
(3)
where Γ(120572) = int
infin
0 119905120572minus1
119890minus119905
119889119905 is the gamma functionThe Riemann-Liouville fractional integral satisfies the
following properties
119868120572
119868120573
119906 (119905) = 119868120573
119868120572
119906 (119905)
119868120572
119868120573
119906 (119905) = 119868120572+120573
119906 (119905)
119868120572
119905119903
=
Γ (119903 + 1)
Γ (120572 + 119903 + 1)
119905120572+119903
(4)
Definition 2 Caputorsquos fractional derivative of order120572 (120572 ge 0)is defined as follows [17]
119863120572
lowast119906 (119905) =
1Γ (119903 minus 120572)
int
119905
0
119906119903
(120591)
(119905 minus 120591)120572minus119903+1 119889120591
0 le 119903 minus 1 lt 120572 lt 119903
(5)
Particularly the operator119863120572lowastsatisfies the following properties
(119888 is a constant)
119863120572
lowast119888 = 0
119863120572
lowast119905120573
=
0 120573 isin N0 120573 lt lceil120572rceil
Γ (120573 + 1)
Γ (120573 + 1 minus 120572)
119905120573minus120572
120573 isin N0 120573 ge lceil120572rceil 120573 notin N 120573 gt lfloor120572rfloor
119863120572
lowast119868120572
119906 (119905) = 119906 (119905)
119868120572
119863120572
lowast119906 (119905) = 119906 (119905) minus
119903minus1
sum
119896=0
119906(119896)
(0+
)
119905119896
119896
119905 gt 0
119863120572
lowast(120582119891 (119905) + 120583119892 (119905)) = 120582119863
120572
lowast(119891 (119905)) + 120583119863
120572
lowast(119892 (119905))
(6)
3 Legendre Polynomials and Some ofTheir Properties
TheLegendre basis polynomials of degree 119899 in [0 1] (see [18])are defined by
119875119894+1 (119905) =
(2119894 + 1) (2119905 minus 1)(119894 + 1)
119875119894(119905) minus
119894
119894 minus 1119875119894minus1 (119905)
119894 = 1 2 (7)
where 1198750(119905) = 1 119875
1(119905) = 2119905minus1 The Legendre polynomials of
degree 119894 can be also written as
119875119894(119905) =
119894
sum
119896=0(minus1)119894+119896 (119894 + 119896)
(119894 minus 119896)
119905119896
(119896)2 (8)
Let
Φ (119905) = [1198750 (119905) 1198751 (119905) 119875119899 (119905)]119879
(9)
The Legendre polynomials given by (7) can be expressed inthe matrix form
Φ (119905) = AT119899(119905) (10)
where
A
=
[
[
[
[
[
[
[
[
[
[
[
[
[
1 0 0 sdot sdot sdot 0
minus1 (minus1)2 2 0 sdot sdot sdot 0
(minus1)2 (minus1)3 31
(minus1)4 42
sdot sdot sdot 0
d
(minus1)119899 (minus1)119899+1 (119899 + 1)(119899 minus 1)
(minus1)119899+1 (119899 + 2)(119899 minus 2)2
sdot sdot sdot (minus1)2119899 (2119899)119899
]
]
]
]
]
]
]
]
]
]
]
]
]
T119899(119905) = [1 119905 119905
119899
]119879
(11)
Obviously
T119899(119905) = Aminus1Φ (119905) (12)
A function 119906(119905) isin 1198712
(0 1) can be expressed in terms of theLegendre basis In practice only the first (119899 + 1) term ofLegendre polynomials is considered Hence
119906 (119905) cong
119899
sum
119894=0119888119894119875119894(119905) = c119879Φ (119905) (13)
where c = [1198880 1198881 119888
119899]119879
119888119894(119894 = 0 1 2 119899) are called
Legendre coefficientsWe extend the notion to two-dimensional space and
define two-dimensional Legendre polynomials of order =
119899 + 1 as a product function of two Legendre polynomials
119875Π(119909 119905) = 119875
119886(119909) 119875119887(119905)
Π = 119886 + 119887 + 1 119886 119887 = 0 1 2 119899(14)
For the function 119906(119909 119905) isin 1198712
([0 1] times [0 1]) we can also getits approximation by using Legendre polynomials
119906 (119909 119905) cong
119899
sum
119894=0
119899
sum
119895=0119906119894119895119875119894(119909) 119875119895(119905) = Φ
119879
(119909)UΦ (119905) (15)
Mathematical Problems in Engineering 3
where
U =
[
[
[
[
[
[
[
[
[
11990600 11990601 sdot sdot sdot 1199060119899
11990610 11990611 sdot sdot sdot 1199061119899
d
1199061198990 1199061198991 sdot sdot sdot 119906
119899119899
]
]
]
]
]
]
]
]
]
(16)
Theorem 3 (see [19]) If a continuous function 119906(119909 119905) definedon [0 1] times [0 1] has bounded mixed fourth partial derivative1205974
119906(119909 119905)1205972
1199091205971199052 then the Legendre expansion of the function
converges uniformly to the functionFor sufficiently smooth function 119906(119909 119905) on [0 1]times[0 1] the
error of the approximation is given by
1003817100381710038171003817119906 (119909 119905) minus 119875
Π(119909 119905)
10038171003817100381710038172 le (1198621 +1198622 +1198623
1+1 )
1+1 (17)
where
1198621=
1
4
max(119909119905)isin[01]times[01]
100381610038161003816100381610038161003816100381610038161003816
120597+1
119906 (119909 119905)
120597119909+1
100381610038161003816100381610038161003816100381610038161003816
1198622=
1
4
max(119909119905)isin[01]times[01]
100381610038161003816100381610038161003816100381610038161003816
120597+1
119906 (119909 119905)
120597119905+1
100381610038161003816100381610038161003816100381610038161003816
1198623=
1
16
max(119909119905)isin[01]times[01]
100381610038161003816100381610038161003816100381610038161003816
1205972+2
119906 (119909 119905)
120597119909+1
120597119905+1
100381610038161003816100381610038161003816100381610038161003816
(18)
We refer the reader to [20] for the proof of the above result
4 Numerical Solution of the Fractional PartialDifferential Equation
Consider the fractional partial differential equationwith vari-able coefficients equation (1) If we approximate the function119906(119909 119905) with the Legendre polynomials it can be written as(15) Then we have
120597120572
119906 (119909 119905)
120597119909120572
≃
120597120572
(Φ119879
(119909)UΦ (119905))
120597119909120572
= (
120597120572
Φ (119909)
120597119909120572
)
119879
UΦ (119905) = (
120597120572T119899(119909)
120597119909120572
)
119879
A119879UΦ (119905)
= [0
Γ (2)
Γ (2 minus 120572)
1199091minus120572
Γ (3)
Γ (3 minus 120572)
1199092minus120572
sdot sdot sdot
Γ (119899 + 1)
Γ (119899 + 1 minus 120572)
119909119899minus120572
]A119879UΦ (119905)
=
[
[
[
[
[
[
[
[
[
[
[
1
119909
1199092
119909119899
]
]
]
]
]
]
]
]
]
]
]
119879
[
[
[
[
[
[
[
[
[
[
[
[
[
[
[
0 0 0 sdot sdot sdot 0
0
Γ (2)
Γ (2 minus 120572)
119909minus120572
0 sdot sdot sdot 0
0 0
Γ (3)
Γ (3 minus 120572)
119909minus120572
sdot sdot sdot 0
d 0
0 0 0 sdot sdot sdot
Γ (119899 + 1)
Γ (119899 + 1 minus 120572)
119909minus120572
]
]
]
]
]
]
]
]
]
]
]
]
]
]
]
A119879UΦ (119905) = Φ119879
(119909) (Aminus1)119879
MA119879UΦ (119905)
120597120573
119906 (119909 119905)
120597119905120573
≃
120597120573
(Φ119879
(119909)UΦ (119905))
120597119905120573
= Φ119879
(119909)U120597120573
Φ (119905)
120597119905120573
= Φ119879
(119909)UA120597120573T119899(119905)
120597119905120573
= Φ119879
(119909)UA [0
Γ (2)
Γ (2 minus 120573)
1199051minus120573
Γ (3)
Γ (3 minus 120573)
1199052minus120573
sdot sdot sdot
Γ (119899 + 1)
Γ (119899 + 1 minus 120573)
119905119899minus120573
]
119879
= Φ119879
(119909)UA
[
[
[
[
[
[
[
[
[
[
[
[
[
[
[
0 0 0 sdot sdot sdot 0
0
Γ (2)
Γ (2 minus 120573)
119905minus120573
0 sdot sdot sdot 0
0 0
Γ (3)
Γ (3 minus 120573)
119905minus120573
sdot sdot sdot 0
d 0
0 0 0 sdot sdot sdot
Γ (119899 + 1)
Γ (119899 + 1 minus 120573)
119905minus120573
]
]
]
]
]
]
]
]
]
]
]
]
]
]
]
T119899(119905) = Φ
119879
(119909)UANAminus1Φ (119905)
(19)
4 Mathematical Problems in Engineering
Let
M
=
[
[
[
[
[
[
[
[
[
[
[
[
[
[
0 0 0 sdot sdot sdot 0
0 Γ (2)Γ (2 minus 120572)
119909minus120572 0 sdot sdot sdot 0
0 0 Γ (3)Γ (3 minus 120572)
119909minus120572
sdot sdot sdot 0
d 0
0 0 0 sdot sdot sdot
Γ (119899 + 1)Γ (119899 + 1 minus 120572)
119909minus120572
]
]
]
]
]
]
]
]
]
]
]
]
]
]
N
=
[
[
[
[
[
[
[
[
[
[
[
[
[
[
[
0 0 0 sdot sdot sdot 0
0
Γ (2)
Γ (2 minus 120573)
119905minus120573
0 sdot sdot sdot 0
0 0
Γ (3)
Γ (3 minus 120573)
119905minus120573
sdot sdot sdot 0
d 0
0 0 0 sdot sdot sdot
Γ (119899 + 1)
Γ (119899 + 1 minus 120573)
119905minus120573
]
]
]
]
]
]
]
]
]
]
]
]
]
]
]
(20)
Substituting (19) into (1) we have
119886 (119909)Φ119879
(119909) (Aminus1)119879
MA119879UΦ (119905)
+ 119887 (119909)Φ119879
(119909)UANAminus1Φ (119905) = 119891 (119909 119905)
(21)
Dispersing (21) by the points (119909119894 119905119895) (119894 = 1 2 119899
119909 119895 =
1 2 119899119905) we can obtain U which is unknown
5 Error Analysis
In this part in order to illustrate the effectiveness of120597120572
119906(119909 119910)120597119909120572
cong Φ119879
(119909)UΦ(119910) we have given the followingtheorem Let 120597120572119906
119899(119909 119910)120597119909
120572 be the following approximationof 120597120572119906(119909 119910)120597119909
120572
120597120572
119906119899(119909 119910)
120597119909120572
=
119899
sum
119894=1
119899
sum
119895=1
119906119894119895119875119894(119909) 119875119895(119910) (22)
Then we have
120597120572
119906 (119909 119910)
120597119909120572
minus
120597120572
119906119899(119909 119910)
120597119909120572
=
infin
sum
119894=119899+1
infin
sum
119895=119899+1119906119894119895119875119894(119909) 119875119895(119910) (23)
Theorem4 Suppose that the function 120597120572
119906119899(119909 119910)120597119909
120572 obtain-ed by using Legendre polynomials is the approximation of120597120572
119906(119909 119910)120597119909120572 and 119906(119909 119910) has bounded mixed fractional
partial derivative |1205974+120572+120573
119906(119909 119910)1205971199092+120572
1205971199102+120573
| le then wehave the following upper bound of error
100381710038171003817100381710038171003817100381710038171003817
120597120572
119906 (119909 119910)
120597119909120572
minus
120597120572
119906119899(119909 119910)
120597119909120572
100381710038171003817100381710038171003817100381710038171003817119864
le
8
(
Γ1015840
(119899 minus 05)
Γ (119899 minus 05)
)
101584010158401015840
(24)
where 119906(119909 119910)119864= (int
1
minus1
int
1
minus1
1199062
(119909 119910)119889119909 119889119910)12 and
119906119894119895
= (
2119894 + 12
)(
2119895 + 12
)
sdot int
1
minus1int
1
minus1
120597120572
119906 (119909 119910)
120597119909120572
119875119894(119909) 119875119895(119910) 119889119909 119889119910
(25)
Proof Theproperty of the sequence 119875119894(119909) on [minus1 1] implies
that
int
1
minus1119875119894(119909) 119875119895(119909) 119889119909 =
22119894 + 1
119894 = 119895
0 119894 = 119895
(26)
then100381710038171003817100381710038171003817100381710038171003817
120597120572
119906 (119909 119910)
120597119909120572
minus
120597120572
119906119899(119909 119910)
120597119909120572
100381710038171003817100381710038171003817100381710038171003817
2
119864
= int
1
minus1int
1
minus1[
120597120572
119906 (119909 119910)
120597119909120572
minus
120597120572
119906119899(119909 119910)
120597119909120572
]
2
119889119909 119889119910
= int
1
minus1int
1
minus1[
[
infin
sum
119894=119899+1
infin
sum
119895=119899+1119906119894119895119875119894(119909) 119875119895(119910)
]
]
2
119889119909 119889119910
= int
1
minus1int
1
minus1
infin
sum
119894=119899+1
infin
sum
119895=119899+111990621198941198951198752119894(119909) 119875
2119895(119910) 119889119909 119889119910
=
infin
sum
119894=119899+1
infin
sum
119895=119899+11199062119894119895int
1
minus11198752119894(119909) 119889119909int
1
minus11198752119895(119910) 119889119910
=
infin
sum
119894=119899+1
infin
sum
119895=119899+11199062119894119895
22119894 + 1
22119895 + 1
(27)
The Legendre polynomials coefficients of function 120597120572
119906(119909 119910)
120597119909120572 are given by
119906119894119895
= (
2119894 + 12
)(
2119895 + 12
)
sdot int
1
minus1int
1
minus1
120597120572
119906 (119909 119910)
120597119909120572
119875119894(119909) 119875119895(119910) 119889119909 119889119910
(28)
Therefore we obtain
119906119894119895
=
2119895 + 14
int
1
minus1
120597120572
119906 (119909 119910)
120597119909120572
[119875119894+1 (119909) minus119875
119894minus1 (119909)]
sdot 119875119895(119910)
100381610038161003816100381610038161003816100381610038161003816
1
minus1119889119910minus
2119895 + 14
sdot int
1
minus1int
1
minus1
120597120572+1
119906 (119909 119910)
120597119909120572+1 [119875
119894+1 (119909) minus 119875119894minus1 (119909)] 119875119895 (119910) 119889119909 119889119910
= minus
2119895 + 14
int
1
minus1int
1
minus1
120597120572+1
119906 (119909 119910)
120597119909120572+1 [119875
119894+1 (119909)
minus 119875119894minus1 (119909)] 119875119895 (119910) 119889119909 119889119910 = minus
2119895 + 14
Mathematical Problems in Engineering 5
sdot int
1
minus1
120597120572+1
119906 (119909 119910)
120597119909120572+1 [
119875119894+2 (119909) minus 119875
119894(119909)
2119894 + 3
minus
119875119894(119909) minus 119875
119894minus2 (119909)
2119894 minus 1]119875119895(119910)
100381610038161003816100381610038161003816100381610038161003816
1
minus1119889119910+
2119895 + 14
sdot int
1
minus1int
1
minus1
120597120572+2
119906 (119909 119910)
120597119909120572+2 [
119875119894+2 (119909) minus 119875
119894(119909)
2119894 + 3
minus
119875119894(119909) minus 119875
119894minus2 (119909)
2119894 minus 1]119875119895(119910) 119889119909 119889119910 =
2119895 + 14
sdot int
1
minus1int
1
minus1
120597120572+2
119906 (119909 119910)
120597119909120572+2 [
119875119894+2 (119909) minus 119875
119894(119909)
2119894 + 3
minus
119875119894(119909) minus 119875
119894minus2 (119909)
2119894 minus 1]119875119895(119910) 119889119909 119889119910
(29)
Now let 120591119894(119909) = (2119894minus1)119875
119894+2(119909)minus2(2119894+1)119875119894(119909)+(2119894+3)119875119894minus2(119909)
then we have
119906119894119895
=
2119895 + 1
4 (2119894 minus 1) (2119894 + 3)
sdot int
1
minus1
int
1
minus1
1205972+120572
119906 (119909 119910)
1205971199092+120572
120591119894(119909) 119875119895(119910) 119889119909 119889119910
(30)
By solving this equation we have
119906119894119895
=
1
4 (2119894 minus 1) (2119894 + 3) (2119895 minus 1) (2119895 + 3)
sdot int
1
minus1
int
1
minus1
1205974+120572+120573
119906 (119909 119910)
1205971199092+120572
1205971199102+120573
120591119894(119909) 120591119895(119910) 119889119909 119889119910
(31)
So we have
10038161003816100381610038161003816119906119894119895
10038161003816100381610038161003816le
14 (2119894 minus 1) (2119894 + 3) (2119895 minus 1) (2119895 + 3)
sdot int
1
minus1int
1
minus1
1003816100381610038161003816100381610038161003816100381610038161003816
1205974+120572+120573
119906 (119909 119910)
1205971199092+120572
1205971199102+120573
1003816100381610038161003816100381610038161003816100381610038161003816
1003816100381610038161003816120591119894(119909)
1003816100381610038161003816
10038161003816100381610038161003816120591119895(119910)
10038161003816100381610038161003816119889119909 119889119910
le
4 (2119894 minus 1) (2119894 + 3) (2119895 minus 1) (2119895 + 3)
sdot int
1
minus1
1003816100381610038161003816120591119894(119909)
1003816100381610038161003816119889119909int
1
minus1
10038161003816100381610038161003816120591119895(119910)
10038161003816100381610038161003816119889119910
(32)
Moreover it was easily obtained that
int
1
minus1
1003816100381610038161003816120591119898
(119905)1003816100381610038161003816119889119905 le radic24 2119894 + 3
radic2119894 minus 3 (33)
thus we have
10038161003816100381610038161003816119906119894119895
10038161003816100381610038161003816le
244 (2119894 minus 1) (2119894 + 3) (2119895 minus 1) (2119895 + 3)
(2119894 + 3)radic2119894 minus 3
sdot
(2119895 + 3)radic2119895 minus 3
le
6(2119894 minus 3)32 (2119895 minus 3)32
(34)
Namely
10038161003816100381610038161003816119906119894119895
10038161003816100381610038161003816
2le
362
(2119894 minus 3)3 (2119895 minus 3)3 (35)
Therefore we have
100381710038171003817100381710038171003817100381710038171003817
120597120572
119906 (119909 119910)
120597119909120572
minus
120597120572
119906119899(119909 119910)
120597119909120572
100381710038171003817100381710038171003817100381710038171003817
2
119864
le
infin
sum
119894=119899+1
infin
sum
119895=119899+11199062119894119895
22119894 + 1
22119895 + 1
le
infin
sum
119894=119899+1
infin
sum
119895=119899+1
1442
(2119894 minus 3)3 (2119895 minus 3)3 (2119894 + 1) (2119895 + 1)
le
infin
sum
119894=119899+1
infin
sum
119895=119899+1
1442
(2119894 minus 3)4 (2119895 minus 3)4
= [
infin
sum
119894=119899+1
12(2119894 minus 3)4
]
2
= [
8(
Γ1015840
(119899 minus 05)Γ (119899 minus 05)
)
101584010158401015840
]
2
(36)
thus
100381710038171003817100381710038171003817100381710038171003817
120597120572
119906 (119909 119910)
120597119909120572
minus
120597120572
119906119899(119909 119910)
120597119909120572
100381710038171003817100381710038171003817100381710038171003817119864
le
8(
Γ1015840
(119899 minus 05)Γ (119899 minus 05)
)
101584010158401015840
(37)
This completes the proof
6 Numerical Examples
Example 1 Consider the following nonhomogeneous partialdifferential equation
11990913 120597
12119906 (119909 119905)
12059711990912 +119909
12 12059712
119906 (119909 119905)
12059711990512 = 119891 (119909 119905)
(119909 119905) isin [0 1] times [0 1] 119906 (119909 0) = 101199092
(1 minus 119909) 119906 (0 119905) = 119906 (1 119905) = 0
(38)
where 119891(119909 119905) = minus40radic119905(3 + 2119905)(minus1 + 119909)11990973
3radic120587 minus 16(1 +
119905)2
1199092
(minus5 + 6119909)3radic120587 The exact solution of this equation is119906(119909 119905) = 10119909
2
(1 minus 119909)(1 + 119905)2 Tables 1ndash3 show the absolute
errors for 119905 = 14119904 119905 = 12119904 and 119905 = 34119904 of different 119899
6 Mathematical Problems in Engineering
Table 1 Absolute error for t = 14 s and different values of 119899
119909 n = 2 n = 3 n = 401 03246 32831e minus 015 33582e minus 01602 02351 42734e minus 015 52745e minus 01503 04820 41237e minus 015 56521e minus 01504 03274 57320e minus 015 62742e minus 01605 08231 47381e minus 015 71640e minus 01606 09127 73722e minus 015 32356e minus 01607 12188 63276e minus 015 42745e minus 01508 15181 12374e minus 014 37224e minus 01509 08364 31744e minus 015 88874e minus 015
Table 2 Absolute error for t = 12 s and different values of 119899
119909 n = 2 n = 3 n = 401 02821 42137e minus 015 34325e minus 01602 04375 58711e minus 015 65332e minus 01603 01021 42210e minus 015 52435e minus 01604 02387 31016e minus 016 49722e minus 01605 08277 34762e minus 016 53478e minus 01506 09322 53265e minus 015 13241e minus 01607 13846 47632e minus 015 92371e minus 01608 16654 42346e minus 016 84812e minus 01509 09144 20437e minus 016 63273e minus 016
Table 3 Absolute error for t = 34 s and different values of 119899
119909 n = 2 n = 3 n = 401 03126 50832e minus 016 53281e minus 01602 08978 41845e minus 016 23258e minus 01503 02374 23448e minus 016 40112e minus 01604 02951 32155e minus 016 51223e minus 01605 03327 55518e minus 015 63274e minus 01506 13267 62440e minus 016 57421e minus 01507 08723 35220e minus 016 52871e minus 01608 09229 42301e minus 016 48810e minus 01609 11327 52310e minus 015 61138e minus 016
From Tables 1ndash3 we can see that the absolute error is verysmall when 119899 ge 3 Also when 119899 is fixed the more pointswe take the more accurate numerical solutions we obtainFigures 1ndash3 show the fact that 119899
119909119905is the number of 119909
119894 119905119895
Example 2 Consider the following fractional partial differ-ential equation
119909
12059714
119906 (119909 119905)
12059711990914 +119909
23 12059713
119906 (119909 119905)
12059711990513 = 119891 (119909 119905)
119906 (119909 0) = 119909 (119909 minus 1) (1199092
+ 1) 119906 (0 119905) = 0
(39)
0
2
4
6
002
0406
081
t0
0204
0608
1
x
u(xt)
Figure 1 Numerical solution of 119899119909119905
= 3
0
2
4
6
002
0406
081
t0
0204
0608
1
x
u(xt)
Figure 2 Numerical solution of 119899119909119905
= 6
0
05
1
00204
06081
0
2
4
6
t
x
u(xt)
Figure 3 Exact solution
where
119891 (119909 119905) =
311990523 (55 + 66119905 + 811199053) (minus1 + 119909) 1199092(1 + 119909
2)
110Γ (23)
+
4 (1 + 119905 + 1199052+ 119905
4) 119909
1712[minus385 + 8119909 (55 minus 60119909 + 641199092
)]
1155Γ (34)
(40)
Mathematical Problems in Engineering 7
0020406081 005
1
0
x t
minus02
minus04
minus06
minus08
minus1
minus12
minus14
u(xt)
Figure 4 The numerical solutions for Example 2 when 119899 = 4
00204060810
051
0
xt
minus02
minus04
minus06
minus08
minus1
minus12
minus14
u(xt)
Figure 5 The numerical solutions for Example 2 when 119899 = 5
The exact solution is 119906(119909 119905) = 119909(119909minus 1)(1199092
+1)(1 + 119905 + 1199052
+ 1199054
)The numerical solutions for 119899 = 4 119899 = 5 are displayed inFigures 4 and 5 and the exact solution is shown in Figure 6
Example 3 Consider this equation
11990912
12059713
119906 (119909 119905)
12059711990913
+11990923
12059713
119906 (119909 119905)
12059711990513
= 119891 (119909 119905)
119906 (119909 0) = 1199092
minus 1 119906 (0 119905) = minus1 minus (119905 minus 1)2
minus (119905 minus 1)3
(41)
where119891 (119909 119905)
=
9 (1 + 119905 minus 21199052 + 1199053) 119909
136
5Γ (23)
+
311990523 (minus2 + 3119905) (minus10 + 9119905) (minus1 + 1199092) 119909
23
40Γ (23)
(42)
The exact solution is 119906(119909 119905) = (1199092
minus 1)[1 + (119905 minus 1)2
+ (119905 minus 1)3
]The numerical solutions for 119899 = 3 119899 = 4 are displayed inFigures 7 and 8 The absolute errors are shown in Figures 9and 10 when 119899 = 3 and 119899 = 4
00204060810
05
1
0
xt
minus02
minus04
minus06
minus08
minus1
minus12
minus14
u(xt)
Figure 6 Exact solution for Example 2
0 02 04 06 08 1
0
05
1
0
u(xt)
x
t
minus02
minus04
minus06
minus08
minus1
minus12
minus14
Figure 7 The numerical solutions for Example 3 when 119899 = 3
0 02 04 06 08 1
0
05
1
0
minus02
minus04
minus06
minus08
minus1
minus12
minus14
u(xt)
xt
Figure 8 The numerical solutions for Example 3 when 119899 = 4
From Examples 1ndash3 we can see that the method in thispaper can be effectively used to solve the numerical solutionof fractional partial differential equation with variable coeffi-cients From the above results the absolute errors between
8 Mathematical Problems in Engineering
002
0406
081 0
05
10020406081
12
tx
times10minus14
u(xt)
Figure 9 The absolute errors for Example 3 when 119899 = 3
002
0406
081 0
0204
0608
10
020406081
12
tx
times10minus15
u(xt)
Figure 10 The absolute errors for Example 3 when 119899 = 4
the numerical solutions and the exact solution are rathersmall What is more due to the absolute error in this paperis about 10minus15 the Legendre polynomials method can reachhigher degree of accuracy by comparing the approximationsobtained by block pulse method [16]
7 Conclusion
In this paper we use the Legendre polynomials method tosolve a class of fractional partial differential equations withvariable coefficients The Legendre polynomials operationalmatrix of fractional differentiation is derived from the prop-erty of Legendre polynomials The initial equation is trans-lated into the product of some relevant matrixes which canalso be regarded as the system of linear equations The erroranalysis of Legendre polynomials is also given The numer-ical results show that numerical solutions obtained by ourmethod are in very good agreement with the exact solution
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
References
[1] R L Bagley and R A Calico ldquoFractional order state equationsfor the control of viscoelastically damped structuresrdquo Journalof Guidance Control and Dynamics vol 14 no 2 pp 304ndash3111991
[2] Z C Li and J S Luo Wavelet Analysis and Its ApplicationElectronic Industrial Publication Beijing China 2005
[3] J H Chen ldquoAnalysis of stability and convergence of numeri-cal approximation for the Riesz fractional reaction-dispersionequationrdquo Journal of Xiamen University (Natural Science) vol46 no 5 pp 616ndash619 2007
[4] D Delbosco and L Rodino ldquoExistence and uniqueness for anonlinear fractional differential equationrdquo Journal of Mathe-matical Analysis and Applications vol 204 no 2 pp 609ndash6251996
[5] Y Ryabov and A Puzenko ldquoA damped oscillations in view ofthe fraction oscillator equationrdquo Physics Review B vol 66 pp184ndash201 2002
[6] A M El-Sayed ldquoNonlinear functional-differential equationsof arbitrary ordersrdquo Nonlinear Analysis Theory Methods ampApplications vol 33 no 2 pp 181ndash186 1998
[7] I L El-Kalla ldquoError estimate of the series solution to a class ofnonlinear fractional differential equationsrdquo Communications inNonlinear Science and Numerical Simulation vol 16 no 3 pp1408ndash1413 2011
[8] Z M Odibat ldquoA study on the convergence of variationaliteration methodrdquo Mathematical and Computer Modelling vol51 no 9-10 pp 1181ndash1192 2010
[9] SMomani ZOdibat andV S Erturk ldquoGeneralized differentialtransform method for solving a space- and time-fractionaldiffusion-wave equationrdquo Physics Letters A vol 370 no 5-6 pp379ndash387 2007
[10] ZOdibat SMomani andV S Erturk ldquoGeneralized differentialtransform method application to differential equations offractional orderrdquo Applied Mathematics and Computation vol197 no 2 pp 467ndash477 2008
[11] Z Odibat and S Momani ldquoA generalized differential transformmethod for linear partial differential equations of fractionalorderrdquo Applied Mathematics Letters vol 21 no 2 pp 194ndash1992008
[12] Y Zhang ldquoA finite difference method for fractional partialdifferential equationrdquo Applied Mathematics and Computationvol 215 no 2 pp 524ndash529 2009
[13] Y X Wang and Q B Fan ldquoThe second kind Chebyshev waveletmethod for solving fractional differential equationsrdquo AppliedMathematics and Computation vol 218 no 17 pp 8592ndash86012012
[14] M X Yi and Y M Chen ldquoHaar wavelet operational matrixmethod for solving fractional partial differential equationsrdquoComputer Modeling in Engineering amp Sciences vol 88 no 3 pp229ndash244 2012
[15] EHDohaAH BhrawyD Baleanu and S S Ezz-Eldien ldquoTheoperational matrix formulation of the Jacobi tau approximationfor space fractional diffusion equationrdquo Advances in DifferenceEquations vol 231 pp 1687ndash1847 2014
[16] M X Yi J Huang and J X Wei ldquoBlock pulse operationalmatrix method for solving fractional partial differential equa-tionrdquo Applied Mathematics and Computation vol 221 pp 121ndash131 2013
Mathematical Problems in Engineering 9
[17] I Podlubny Fractional Differential Equations vol 198 ofMath-ematics in Science and Engineering Academic Press New YorkNY USA 1999
[18] A Saadatmandi and M Dehghan ldquoA new operational matrixfor solving fractional-order differential equationsrdquo ComputersandMathematics with Applications vol 59 no 3 pp 1326ndash13362010
[19] N Liu and E-B Lin ldquoLegendre wavelet method for numericalsolutions of partial differential equationsrdquo Numerical Methodsfor Partial Differential Equations vol 26 no 1 pp 81ndash94 2010
[20] S Nemati and Y Ordokhani ldquoLegendre expansion methodsfor the numerical solution of nonlinear 2D Fredholm integralequations of the second kindrdquo Journal of Applied Mathematicsamp Informatics vol 31 no 5-6 pp 609ndash621 2013
Submit your manuscripts athttpwwwhindawicom
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Mathematical Problems in Engineering
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Differential EquationsInternational Journal of
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Mathematical PhysicsAdvances in
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OptimizationJournal of
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International Journal of
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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
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Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
2 Mathematical Problems in Engineering
2 Definitions of FractionalDerivatives and Integrals
Definition 1 Riemann-Liouville fractional integral of order120572 (120572 ge 0) is defined as follows [17]
119868120572
119906 (119905) =
1Γ (120572)
int
119905
0(119905 minus 120591)
120572minus1119906 (120591) 119889120591 119905 gt 0
1198680119906 (119905) = 119906 (119905)
(3)
where Γ(120572) = int
infin
0 119905120572minus1
119890minus119905
119889119905 is the gamma functionThe Riemann-Liouville fractional integral satisfies the
following properties
119868120572
119868120573
119906 (119905) = 119868120573
119868120572
119906 (119905)
119868120572
119868120573
119906 (119905) = 119868120572+120573
119906 (119905)
119868120572
119905119903
=
Γ (119903 + 1)
Γ (120572 + 119903 + 1)
119905120572+119903
(4)
Definition 2 Caputorsquos fractional derivative of order120572 (120572 ge 0)is defined as follows [17]
119863120572
lowast119906 (119905) =
1Γ (119903 minus 120572)
int
119905
0
119906119903
(120591)
(119905 minus 120591)120572minus119903+1 119889120591
0 le 119903 minus 1 lt 120572 lt 119903
(5)
Particularly the operator119863120572lowastsatisfies the following properties
(119888 is a constant)
119863120572
lowast119888 = 0
119863120572
lowast119905120573
=
0 120573 isin N0 120573 lt lceil120572rceil
Γ (120573 + 1)
Γ (120573 + 1 minus 120572)
119905120573minus120572
120573 isin N0 120573 ge lceil120572rceil 120573 notin N 120573 gt lfloor120572rfloor
119863120572
lowast119868120572
119906 (119905) = 119906 (119905)
119868120572
119863120572
lowast119906 (119905) = 119906 (119905) minus
119903minus1
sum
119896=0
119906(119896)
(0+
)
119905119896
119896
119905 gt 0
119863120572
lowast(120582119891 (119905) + 120583119892 (119905)) = 120582119863
120572
lowast(119891 (119905)) + 120583119863
120572
lowast(119892 (119905))
(6)
3 Legendre Polynomials and Some ofTheir Properties
TheLegendre basis polynomials of degree 119899 in [0 1] (see [18])are defined by
119875119894+1 (119905) =
(2119894 + 1) (2119905 minus 1)(119894 + 1)
119875119894(119905) minus
119894
119894 minus 1119875119894minus1 (119905)
119894 = 1 2 (7)
where 1198750(119905) = 1 119875
1(119905) = 2119905minus1 The Legendre polynomials of
degree 119894 can be also written as
119875119894(119905) =
119894
sum
119896=0(minus1)119894+119896 (119894 + 119896)
(119894 minus 119896)
119905119896
(119896)2 (8)
Let
Φ (119905) = [1198750 (119905) 1198751 (119905) 119875119899 (119905)]119879
(9)
The Legendre polynomials given by (7) can be expressed inthe matrix form
Φ (119905) = AT119899(119905) (10)
where
A
=
[
[
[
[
[
[
[
[
[
[
[
[
[
1 0 0 sdot sdot sdot 0
minus1 (minus1)2 2 0 sdot sdot sdot 0
(minus1)2 (minus1)3 31
(minus1)4 42
sdot sdot sdot 0
d
(minus1)119899 (minus1)119899+1 (119899 + 1)(119899 minus 1)
(minus1)119899+1 (119899 + 2)(119899 minus 2)2
sdot sdot sdot (minus1)2119899 (2119899)119899
]
]
]
]
]
]
]
]
]
]
]
]
]
T119899(119905) = [1 119905 119905
119899
]119879
(11)
Obviously
T119899(119905) = Aminus1Φ (119905) (12)
A function 119906(119905) isin 1198712
(0 1) can be expressed in terms of theLegendre basis In practice only the first (119899 + 1) term ofLegendre polynomials is considered Hence
119906 (119905) cong
119899
sum
119894=0119888119894119875119894(119905) = c119879Φ (119905) (13)
where c = [1198880 1198881 119888
119899]119879
119888119894(119894 = 0 1 2 119899) are called
Legendre coefficientsWe extend the notion to two-dimensional space and
define two-dimensional Legendre polynomials of order =
119899 + 1 as a product function of two Legendre polynomials
119875Π(119909 119905) = 119875
119886(119909) 119875119887(119905)
Π = 119886 + 119887 + 1 119886 119887 = 0 1 2 119899(14)
For the function 119906(119909 119905) isin 1198712
([0 1] times [0 1]) we can also getits approximation by using Legendre polynomials
119906 (119909 119905) cong
119899
sum
119894=0
119899
sum
119895=0119906119894119895119875119894(119909) 119875119895(119905) = Φ
119879
(119909)UΦ (119905) (15)
Mathematical Problems in Engineering 3
where
U =
[
[
[
[
[
[
[
[
[
11990600 11990601 sdot sdot sdot 1199060119899
11990610 11990611 sdot sdot sdot 1199061119899
d
1199061198990 1199061198991 sdot sdot sdot 119906
119899119899
]
]
]
]
]
]
]
]
]
(16)
Theorem 3 (see [19]) If a continuous function 119906(119909 119905) definedon [0 1] times [0 1] has bounded mixed fourth partial derivative1205974
119906(119909 119905)1205972
1199091205971199052 then the Legendre expansion of the function
converges uniformly to the functionFor sufficiently smooth function 119906(119909 119905) on [0 1]times[0 1] the
error of the approximation is given by
1003817100381710038171003817119906 (119909 119905) minus 119875
Π(119909 119905)
10038171003817100381710038172 le (1198621 +1198622 +1198623
1+1 )
1+1 (17)
where
1198621=
1
4
max(119909119905)isin[01]times[01]
100381610038161003816100381610038161003816100381610038161003816
120597+1
119906 (119909 119905)
120597119909+1
100381610038161003816100381610038161003816100381610038161003816
1198622=
1
4
max(119909119905)isin[01]times[01]
100381610038161003816100381610038161003816100381610038161003816
120597+1
119906 (119909 119905)
120597119905+1
100381610038161003816100381610038161003816100381610038161003816
1198623=
1
16
max(119909119905)isin[01]times[01]
100381610038161003816100381610038161003816100381610038161003816
1205972+2
119906 (119909 119905)
120597119909+1
120597119905+1
100381610038161003816100381610038161003816100381610038161003816
(18)
We refer the reader to [20] for the proof of the above result
4 Numerical Solution of the Fractional PartialDifferential Equation
Consider the fractional partial differential equationwith vari-able coefficients equation (1) If we approximate the function119906(119909 119905) with the Legendre polynomials it can be written as(15) Then we have
120597120572
119906 (119909 119905)
120597119909120572
≃
120597120572
(Φ119879
(119909)UΦ (119905))
120597119909120572
= (
120597120572
Φ (119909)
120597119909120572
)
119879
UΦ (119905) = (
120597120572T119899(119909)
120597119909120572
)
119879
A119879UΦ (119905)
= [0
Γ (2)
Γ (2 minus 120572)
1199091minus120572
Γ (3)
Γ (3 minus 120572)
1199092minus120572
sdot sdot sdot
Γ (119899 + 1)
Γ (119899 + 1 minus 120572)
119909119899minus120572
]A119879UΦ (119905)
=
[
[
[
[
[
[
[
[
[
[
[
1
119909
1199092
119909119899
]
]
]
]
]
]
]
]
]
]
]
119879
[
[
[
[
[
[
[
[
[
[
[
[
[
[
[
0 0 0 sdot sdot sdot 0
0
Γ (2)
Γ (2 minus 120572)
119909minus120572
0 sdot sdot sdot 0
0 0
Γ (3)
Γ (3 minus 120572)
119909minus120572
sdot sdot sdot 0
d 0
0 0 0 sdot sdot sdot
Γ (119899 + 1)
Γ (119899 + 1 minus 120572)
119909minus120572
]
]
]
]
]
]
]
]
]
]
]
]
]
]
]
A119879UΦ (119905) = Φ119879
(119909) (Aminus1)119879
MA119879UΦ (119905)
120597120573
119906 (119909 119905)
120597119905120573
≃
120597120573
(Φ119879
(119909)UΦ (119905))
120597119905120573
= Φ119879
(119909)U120597120573
Φ (119905)
120597119905120573
= Φ119879
(119909)UA120597120573T119899(119905)
120597119905120573
= Φ119879
(119909)UA [0
Γ (2)
Γ (2 minus 120573)
1199051minus120573
Γ (3)
Γ (3 minus 120573)
1199052minus120573
sdot sdot sdot
Γ (119899 + 1)
Γ (119899 + 1 minus 120573)
119905119899minus120573
]
119879
= Φ119879
(119909)UA
[
[
[
[
[
[
[
[
[
[
[
[
[
[
[
0 0 0 sdot sdot sdot 0
0
Γ (2)
Γ (2 minus 120573)
119905minus120573
0 sdot sdot sdot 0
0 0
Γ (3)
Γ (3 minus 120573)
119905minus120573
sdot sdot sdot 0
d 0
0 0 0 sdot sdot sdot
Γ (119899 + 1)
Γ (119899 + 1 minus 120573)
119905minus120573
]
]
]
]
]
]
]
]
]
]
]
]
]
]
]
T119899(119905) = Φ
119879
(119909)UANAminus1Φ (119905)
(19)
4 Mathematical Problems in Engineering
Let
M
=
[
[
[
[
[
[
[
[
[
[
[
[
[
[
0 0 0 sdot sdot sdot 0
0 Γ (2)Γ (2 minus 120572)
119909minus120572 0 sdot sdot sdot 0
0 0 Γ (3)Γ (3 minus 120572)
119909minus120572
sdot sdot sdot 0
d 0
0 0 0 sdot sdot sdot
Γ (119899 + 1)Γ (119899 + 1 minus 120572)
119909minus120572
]
]
]
]
]
]
]
]
]
]
]
]
]
]
N
=
[
[
[
[
[
[
[
[
[
[
[
[
[
[
[
0 0 0 sdot sdot sdot 0
0
Γ (2)
Γ (2 minus 120573)
119905minus120573
0 sdot sdot sdot 0
0 0
Γ (3)
Γ (3 minus 120573)
119905minus120573
sdot sdot sdot 0
d 0
0 0 0 sdot sdot sdot
Γ (119899 + 1)
Γ (119899 + 1 minus 120573)
119905minus120573
]
]
]
]
]
]
]
]
]
]
]
]
]
]
]
(20)
Substituting (19) into (1) we have
119886 (119909)Φ119879
(119909) (Aminus1)119879
MA119879UΦ (119905)
+ 119887 (119909)Φ119879
(119909)UANAminus1Φ (119905) = 119891 (119909 119905)
(21)
Dispersing (21) by the points (119909119894 119905119895) (119894 = 1 2 119899
119909 119895 =
1 2 119899119905) we can obtain U which is unknown
5 Error Analysis
In this part in order to illustrate the effectiveness of120597120572
119906(119909 119910)120597119909120572
cong Φ119879
(119909)UΦ(119910) we have given the followingtheorem Let 120597120572119906
119899(119909 119910)120597119909
120572 be the following approximationof 120597120572119906(119909 119910)120597119909
120572
120597120572
119906119899(119909 119910)
120597119909120572
=
119899
sum
119894=1
119899
sum
119895=1
119906119894119895119875119894(119909) 119875119895(119910) (22)
Then we have
120597120572
119906 (119909 119910)
120597119909120572
minus
120597120572
119906119899(119909 119910)
120597119909120572
=
infin
sum
119894=119899+1
infin
sum
119895=119899+1119906119894119895119875119894(119909) 119875119895(119910) (23)
Theorem4 Suppose that the function 120597120572
119906119899(119909 119910)120597119909
120572 obtain-ed by using Legendre polynomials is the approximation of120597120572
119906(119909 119910)120597119909120572 and 119906(119909 119910) has bounded mixed fractional
partial derivative |1205974+120572+120573
119906(119909 119910)1205971199092+120572
1205971199102+120573
| le then wehave the following upper bound of error
100381710038171003817100381710038171003817100381710038171003817
120597120572
119906 (119909 119910)
120597119909120572
minus
120597120572
119906119899(119909 119910)
120597119909120572
100381710038171003817100381710038171003817100381710038171003817119864
le
8
(
Γ1015840
(119899 minus 05)
Γ (119899 minus 05)
)
101584010158401015840
(24)
where 119906(119909 119910)119864= (int
1
minus1
int
1
minus1
1199062
(119909 119910)119889119909 119889119910)12 and
119906119894119895
= (
2119894 + 12
)(
2119895 + 12
)
sdot int
1
minus1int
1
minus1
120597120572
119906 (119909 119910)
120597119909120572
119875119894(119909) 119875119895(119910) 119889119909 119889119910
(25)
Proof Theproperty of the sequence 119875119894(119909) on [minus1 1] implies
that
int
1
minus1119875119894(119909) 119875119895(119909) 119889119909 =
22119894 + 1
119894 = 119895
0 119894 = 119895
(26)
then100381710038171003817100381710038171003817100381710038171003817
120597120572
119906 (119909 119910)
120597119909120572
minus
120597120572
119906119899(119909 119910)
120597119909120572
100381710038171003817100381710038171003817100381710038171003817
2
119864
= int
1
minus1int
1
minus1[
120597120572
119906 (119909 119910)
120597119909120572
minus
120597120572
119906119899(119909 119910)
120597119909120572
]
2
119889119909 119889119910
= int
1
minus1int
1
minus1[
[
infin
sum
119894=119899+1
infin
sum
119895=119899+1119906119894119895119875119894(119909) 119875119895(119910)
]
]
2
119889119909 119889119910
= int
1
minus1int
1
minus1
infin
sum
119894=119899+1
infin
sum
119895=119899+111990621198941198951198752119894(119909) 119875
2119895(119910) 119889119909 119889119910
=
infin
sum
119894=119899+1
infin
sum
119895=119899+11199062119894119895int
1
minus11198752119894(119909) 119889119909int
1
minus11198752119895(119910) 119889119910
=
infin
sum
119894=119899+1
infin
sum
119895=119899+11199062119894119895
22119894 + 1
22119895 + 1
(27)
The Legendre polynomials coefficients of function 120597120572
119906(119909 119910)
120597119909120572 are given by
119906119894119895
= (
2119894 + 12
)(
2119895 + 12
)
sdot int
1
minus1int
1
minus1
120597120572
119906 (119909 119910)
120597119909120572
119875119894(119909) 119875119895(119910) 119889119909 119889119910
(28)
Therefore we obtain
119906119894119895
=
2119895 + 14
int
1
minus1
120597120572
119906 (119909 119910)
120597119909120572
[119875119894+1 (119909) minus119875
119894minus1 (119909)]
sdot 119875119895(119910)
100381610038161003816100381610038161003816100381610038161003816
1
minus1119889119910minus
2119895 + 14
sdot int
1
minus1int
1
minus1
120597120572+1
119906 (119909 119910)
120597119909120572+1 [119875
119894+1 (119909) minus 119875119894minus1 (119909)] 119875119895 (119910) 119889119909 119889119910
= minus
2119895 + 14
int
1
minus1int
1
minus1
120597120572+1
119906 (119909 119910)
120597119909120572+1 [119875
119894+1 (119909)
minus 119875119894minus1 (119909)] 119875119895 (119910) 119889119909 119889119910 = minus
2119895 + 14
Mathematical Problems in Engineering 5
sdot int
1
minus1
120597120572+1
119906 (119909 119910)
120597119909120572+1 [
119875119894+2 (119909) minus 119875
119894(119909)
2119894 + 3
minus
119875119894(119909) minus 119875
119894minus2 (119909)
2119894 minus 1]119875119895(119910)
100381610038161003816100381610038161003816100381610038161003816
1
minus1119889119910+
2119895 + 14
sdot int
1
minus1int
1
minus1
120597120572+2
119906 (119909 119910)
120597119909120572+2 [
119875119894+2 (119909) minus 119875
119894(119909)
2119894 + 3
minus
119875119894(119909) minus 119875
119894minus2 (119909)
2119894 minus 1]119875119895(119910) 119889119909 119889119910 =
2119895 + 14
sdot int
1
minus1int
1
minus1
120597120572+2
119906 (119909 119910)
120597119909120572+2 [
119875119894+2 (119909) minus 119875
119894(119909)
2119894 + 3
minus
119875119894(119909) minus 119875
119894minus2 (119909)
2119894 minus 1]119875119895(119910) 119889119909 119889119910
(29)
Now let 120591119894(119909) = (2119894minus1)119875
119894+2(119909)minus2(2119894+1)119875119894(119909)+(2119894+3)119875119894minus2(119909)
then we have
119906119894119895
=
2119895 + 1
4 (2119894 minus 1) (2119894 + 3)
sdot int
1
minus1
int
1
minus1
1205972+120572
119906 (119909 119910)
1205971199092+120572
120591119894(119909) 119875119895(119910) 119889119909 119889119910
(30)
By solving this equation we have
119906119894119895
=
1
4 (2119894 minus 1) (2119894 + 3) (2119895 minus 1) (2119895 + 3)
sdot int
1
minus1
int
1
minus1
1205974+120572+120573
119906 (119909 119910)
1205971199092+120572
1205971199102+120573
120591119894(119909) 120591119895(119910) 119889119909 119889119910
(31)
So we have
10038161003816100381610038161003816119906119894119895
10038161003816100381610038161003816le
14 (2119894 minus 1) (2119894 + 3) (2119895 minus 1) (2119895 + 3)
sdot int
1
minus1int
1
minus1
1003816100381610038161003816100381610038161003816100381610038161003816
1205974+120572+120573
119906 (119909 119910)
1205971199092+120572
1205971199102+120573
1003816100381610038161003816100381610038161003816100381610038161003816
1003816100381610038161003816120591119894(119909)
1003816100381610038161003816
10038161003816100381610038161003816120591119895(119910)
10038161003816100381610038161003816119889119909 119889119910
le
4 (2119894 minus 1) (2119894 + 3) (2119895 minus 1) (2119895 + 3)
sdot int
1
minus1
1003816100381610038161003816120591119894(119909)
1003816100381610038161003816119889119909int
1
minus1
10038161003816100381610038161003816120591119895(119910)
10038161003816100381610038161003816119889119910
(32)
Moreover it was easily obtained that
int
1
minus1
1003816100381610038161003816120591119898
(119905)1003816100381610038161003816119889119905 le radic24 2119894 + 3
radic2119894 minus 3 (33)
thus we have
10038161003816100381610038161003816119906119894119895
10038161003816100381610038161003816le
244 (2119894 minus 1) (2119894 + 3) (2119895 minus 1) (2119895 + 3)
(2119894 + 3)radic2119894 minus 3
sdot
(2119895 + 3)radic2119895 minus 3
le
6(2119894 minus 3)32 (2119895 minus 3)32
(34)
Namely
10038161003816100381610038161003816119906119894119895
10038161003816100381610038161003816
2le
362
(2119894 minus 3)3 (2119895 minus 3)3 (35)
Therefore we have
100381710038171003817100381710038171003817100381710038171003817
120597120572
119906 (119909 119910)
120597119909120572
minus
120597120572
119906119899(119909 119910)
120597119909120572
100381710038171003817100381710038171003817100381710038171003817
2
119864
le
infin
sum
119894=119899+1
infin
sum
119895=119899+11199062119894119895
22119894 + 1
22119895 + 1
le
infin
sum
119894=119899+1
infin
sum
119895=119899+1
1442
(2119894 minus 3)3 (2119895 minus 3)3 (2119894 + 1) (2119895 + 1)
le
infin
sum
119894=119899+1
infin
sum
119895=119899+1
1442
(2119894 minus 3)4 (2119895 minus 3)4
= [
infin
sum
119894=119899+1
12(2119894 minus 3)4
]
2
= [
8(
Γ1015840
(119899 minus 05)Γ (119899 minus 05)
)
101584010158401015840
]
2
(36)
thus
100381710038171003817100381710038171003817100381710038171003817
120597120572
119906 (119909 119910)
120597119909120572
minus
120597120572
119906119899(119909 119910)
120597119909120572
100381710038171003817100381710038171003817100381710038171003817119864
le
8(
Γ1015840
(119899 minus 05)Γ (119899 minus 05)
)
101584010158401015840
(37)
This completes the proof
6 Numerical Examples
Example 1 Consider the following nonhomogeneous partialdifferential equation
11990913 120597
12119906 (119909 119905)
12059711990912 +119909
12 12059712
119906 (119909 119905)
12059711990512 = 119891 (119909 119905)
(119909 119905) isin [0 1] times [0 1] 119906 (119909 0) = 101199092
(1 minus 119909) 119906 (0 119905) = 119906 (1 119905) = 0
(38)
where 119891(119909 119905) = minus40radic119905(3 + 2119905)(minus1 + 119909)11990973
3radic120587 minus 16(1 +
119905)2
1199092
(minus5 + 6119909)3radic120587 The exact solution of this equation is119906(119909 119905) = 10119909
2
(1 minus 119909)(1 + 119905)2 Tables 1ndash3 show the absolute
errors for 119905 = 14119904 119905 = 12119904 and 119905 = 34119904 of different 119899
6 Mathematical Problems in Engineering
Table 1 Absolute error for t = 14 s and different values of 119899
119909 n = 2 n = 3 n = 401 03246 32831e minus 015 33582e minus 01602 02351 42734e minus 015 52745e minus 01503 04820 41237e minus 015 56521e minus 01504 03274 57320e minus 015 62742e minus 01605 08231 47381e minus 015 71640e minus 01606 09127 73722e minus 015 32356e minus 01607 12188 63276e minus 015 42745e minus 01508 15181 12374e minus 014 37224e minus 01509 08364 31744e minus 015 88874e minus 015
Table 2 Absolute error for t = 12 s and different values of 119899
119909 n = 2 n = 3 n = 401 02821 42137e minus 015 34325e minus 01602 04375 58711e minus 015 65332e minus 01603 01021 42210e minus 015 52435e minus 01604 02387 31016e minus 016 49722e minus 01605 08277 34762e minus 016 53478e minus 01506 09322 53265e minus 015 13241e minus 01607 13846 47632e minus 015 92371e minus 01608 16654 42346e minus 016 84812e minus 01509 09144 20437e minus 016 63273e minus 016
Table 3 Absolute error for t = 34 s and different values of 119899
119909 n = 2 n = 3 n = 401 03126 50832e minus 016 53281e minus 01602 08978 41845e minus 016 23258e minus 01503 02374 23448e minus 016 40112e minus 01604 02951 32155e minus 016 51223e minus 01605 03327 55518e minus 015 63274e minus 01506 13267 62440e minus 016 57421e minus 01507 08723 35220e minus 016 52871e minus 01608 09229 42301e minus 016 48810e minus 01609 11327 52310e minus 015 61138e minus 016
From Tables 1ndash3 we can see that the absolute error is verysmall when 119899 ge 3 Also when 119899 is fixed the more pointswe take the more accurate numerical solutions we obtainFigures 1ndash3 show the fact that 119899
119909119905is the number of 119909
119894 119905119895
Example 2 Consider the following fractional partial differ-ential equation
119909
12059714
119906 (119909 119905)
12059711990914 +119909
23 12059713
119906 (119909 119905)
12059711990513 = 119891 (119909 119905)
119906 (119909 0) = 119909 (119909 minus 1) (1199092
+ 1) 119906 (0 119905) = 0
(39)
0
2
4
6
002
0406
081
t0
0204
0608
1
x
u(xt)
Figure 1 Numerical solution of 119899119909119905
= 3
0
2
4
6
002
0406
081
t0
0204
0608
1
x
u(xt)
Figure 2 Numerical solution of 119899119909119905
= 6
0
05
1
00204
06081
0
2
4
6
t
x
u(xt)
Figure 3 Exact solution
where
119891 (119909 119905) =
311990523 (55 + 66119905 + 811199053) (minus1 + 119909) 1199092(1 + 119909
2)
110Γ (23)
+
4 (1 + 119905 + 1199052+ 119905
4) 119909
1712[minus385 + 8119909 (55 minus 60119909 + 641199092
)]
1155Γ (34)
(40)
Mathematical Problems in Engineering 7
0020406081 005
1
0
x t
minus02
minus04
minus06
minus08
minus1
minus12
minus14
u(xt)
Figure 4 The numerical solutions for Example 2 when 119899 = 4
00204060810
051
0
xt
minus02
minus04
minus06
minus08
minus1
minus12
minus14
u(xt)
Figure 5 The numerical solutions for Example 2 when 119899 = 5
The exact solution is 119906(119909 119905) = 119909(119909minus 1)(1199092
+1)(1 + 119905 + 1199052
+ 1199054
)The numerical solutions for 119899 = 4 119899 = 5 are displayed inFigures 4 and 5 and the exact solution is shown in Figure 6
Example 3 Consider this equation
11990912
12059713
119906 (119909 119905)
12059711990913
+11990923
12059713
119906 (119909 119905)
12059711990513
= 119891 (119909 119905)
119906 (119909 0) = 1199092
minus 1 119906 (0 119905) = minus1 minus (119905 minus 1)2
minus (119905 minus 1)3
(41)
where119891 (119909 119905)
=
9 (1 + 119905 minus 21199052 + 1199053) 119909
136
5Γ (23)
+
311990523 (minus2 + 3119905) (minus10 + 9119905) (minus1 + 1199092) 119909
23
40Γ (23)
(42)
The exact solution is 119906(119909 119905) = (1199092
minus 1)[1 + (119905 minus 1)2
+ (119905 minus 1)3
]The numerical solutions for 119899 = 3 119899 = 4 are displayed inFigures 7 and 8 The absolute errors are shown in Figures 9and 10 when 119899 = 3 and 119899 = 4
00204060810
05
1
0
xt
minus02
minus04
minus06
minus08
minus1
minus12
minus14
u(xt)
Figure 6 Exact solution for Example 2
0 02 04 06 08 1
0
05
1
0
u(xt)
x
t
minus02
minus04
minus06
minus08
minus1
minus12
minus14
Figure 7 The numerical solutions for Example 3 when 119899 = 3
0 02 04 06 08 1
0
05
1
0
minus02
minus04
minus06
minus08
minus1
minus12
minus14
u(xt)
xt
Figure 8 The numerical solutions for Example 3 when 119899 = 4
From Examples 1ndash3 we can see that the method in thispaper can be effectively used to solve the numerical solutionof fractional partial differential equation with variable coeffi-cients From the above results the absolute errors between
8 Mathematical Problems in Engineering
002
0406
081 0
05
10020406081
12
tx
times10minus14
u(xt)
Figure 9 The absolute errors for Example 3 when 119899 = 3
002
0406
081 0
0204
0608
10
020406081
12
tx
times10minus15
u(xt)
Figure 10 The absolute errors for Example 3 when 119899 = 4
the numerical solutions and the exact solution are rathersmall What is more due to the absolute error in this paperis about 10minus15 the Legendre polynomials method can reachhigher degree of accuracy by comparing the approximationsobtained by block pulse method [16]
7 Conclusion
In this paper we use the Legendre polynomials method tosolve a class of fractional partial differential equations withvariable coefficients The Legendre polynomials operationalmatrix of fractional differentiation is derived from the prop-erty of Legendre polynomials The initial equation is trans-lated into the product of some relevant matrixes which canalso be regarded as the system of linear equations The erroranalysis of Legendre polynomials is also given The numer-ical results show that numerical solutions obtained by ourmethod are in very good agreement with the exact solution
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
References
[1] R L Bagley and R A Calico ldquoFractional order state equationsfor the control of viscoelastically damped structuresrdquo Journalof Guidance Control and Dynamics vol 14 no 2 pp 304ndash3111991
[2] Z C Li and J S Luo Wavelet Analysis and Its ApplicationElectronic Industrial Publication Beijing China 2005
[3] J H Chen ldquoAnalysis of stability and convergence of numeri-cal approximation for the Riesz fractional reaction-dispersionequationrdquo Journal of Xiamen University (Natural Science) vol46 no 5 pp 616ndash619 2007
[4] D Delbosco and L Rodino ldquoExistence and uniqueness for anonlinear fractional differential equationrdquo Journal of Mathe-matical Analysis and Applications vol 204 no 2 pp 609ndash6251996
[5] Y Ryabov and A Puzenko ldquoA damped oscillations in view ofthe fraction oscillator equationrdquo Physics Review B vol 66 pp184ndash201 2002
[6] A M El-Sayed ldquoNonlinear functional-differential equationsof arbitrary ordersrdquo Nonlinear Analysis Theory Methods ampApplications vol 33 no 2 pp 181ndash186 1998
[7] I L El-Kalla ldquoError estimate of the series solution to a class ofnonlinear fractional differential equationsrdquo Communications inNonlinear Science and Numerical Simulation vol 16 no 3 pp1408ndash1413 2011
[8] Z M Odibat ldquoA study on the convergence of variationaliteration methodrdquo Mathematical and Computer Modelling vol51 no 9-10 pp 1181ndash1192 2010
[9] SMomani ZOdibat andV S Erturk ldquoGeneralized differentialtransform method for solving a space- and time-fractionaldiffusion-wave equationrdquo Physics Letters A vol 370 no 5-6 pp379ndash387 2007
[10] ZOdibat SMomani andV S Erturk ldquoGeneralized differentialtransform method application to differential equations offractional orderrdquo Applied Mathematics and Computation vol197 no 2 pp 467ndash477 2008
[11] Z Odibat and S Momani ldquoA generalized differential transformmethod for linear partial differential equations of fractionalorderrdquo Applied Mathematics Letters vol 21 no 2 pp 194ndash1992008
[12] Y Zhang ldquoA finite difference method for fractional partialdifferential equationrdquo Applied Mathematics and Computationvol 215 no 2 pp 524ndash529 2009
[13] Y X Wang and Q B Fan ldquoThe second kind Chebyshev waveletmethod for solving fractional differential equationsrdquo AppliedMathematics and Computation vol 218 no 17 pp 8592ndash86012012
[14] M X Yi and Y M Chen ldquoHaar wavelet operational matrixmethod for solving fractional partial differential equationsrdquoComputer Modeling in Engineering amp Sciences vol 88 no 3 pp229ndash244 2012
[15] EHDohaAH BhrawyD Baleanu and S S Ezz-Eldien ldquoTheoperational matrix formulation of the Jacobi tau approximationfor space fractional diffusion equationrdquo Advances in DifferenceEquations vol 231 pp 1687ndash1847 2014
[16] M X Yi J Huang and J X Wei ldquoBlock pulse operationalmatrix method for solving fractional partial differential equa-tionrdquo Applied Mathematics and Computation vol 221 pp 121ndash131 2013
Mathematical Problems in Engineering 9
[17] I Podlubny Fractional Differential Equations vol 198 ofMath-ematics in Science and Engineering Academic Press New YorkNY USA 1999
[18] A Saadatmandi and M Dehghan ldquoA new operational matrixfor solving fractional-order differential equationsrdquo ComputersandMathematics with Applications vol 59 no 3 pp 1326ndash13362010
[19] N Liu and E-B Lin ldquoLegendre wavelet method for numericalsolutions of partial differential equationsrdquo Numerical Methodsfor Partial Differential Equations vol 26 no 1 pp 81ndash94 2010
[20] S Nemati and Y Ordokhani ldquoLegendre expansion methodsfor the numerical solution of nonlinear 2D Fredholm integralequations of the second kindrdquo Journal of Applied Mathematicsamp Informatics vol 31 no 5-6 pp 609ndash621 2013
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
Mathematical Problems in Engineering 3
where
U =
[
[
[
[
[
[
[
[
[
11990600 11990601 sdot sdot sdot 1199060119899
11990610 11990611 sdot sdot sdot 1199061119899
d
1199061198990 1199061198991 sdot sdot sdot 119906
119899119899
]
]
]
]
]
]
]
]
]
(16)
Theorem 3 (see [19]) If a continuous function 119906(119909 119905) definedon [0 1] times [0 1] has bounded mixed fourth partial derivative1205974
119906(119909 119905)1205972
1199091205971199052 then the Legendre expansion of the function
converges uniformly to the functionFor sufficiently smooth function 119906(119909 119905) on [0 1]times[0 1] the
error of the approximation is given by
1003817100381710038171003817119906 (119909 119905) minus 119875
Π(119909 119905)
10038171003817100381710038172 le (1198621 +1198622 +1198623
1+1 )
1+1 (17)
where
1198621=
1
4
max(119909119905)isin[01]times[01]
100381610038161003816100381610038161003816100381610038161003816
120597+1
119906 (119909 119905)
120597119909+1
100381610038161003816100381610038161003816100381610038161003816
1198622=
1
4
max(119909119905)isin[01]times[01]
100381610038161003816100381610038161003816100381610038161003816
120597+1
119906 (119909 119905)
120597119905+1
100381610038161003816100381610038161003816100381610038161003816
1198623=
1
16
max(119909119905)isin[01]times[01]
100381610038161003816100381610038161003816100381610038161003816
1205972+2
119906 (119909 119905)
120597119909+1
120597119905+1
100381610038161003816100381610038161003816100381610038161003816
(18)
We refer the reader to [20] for the proof of the above result
4 Numerical Solution of the Fractional PartialDifferential Equation
Consider the fractional partial differential equationwith vari-able coefficients equation (1) If we approximate the function119906(119909 119905) with the Legendre polynomials it can be written as(15) Then we have
120597120572
119906 (119909 119905)
120597119909120572
≃
120597120572
(Φ119879
(119909)UΦ (119905))
120597119909120572
= (
120597120572
Φ (119909)
120597119909120572
)
119879
UΦ (119905) = (
120597120572T119899(119909)
120597119909120572
)
119879
A119879UΦ (119905)
= [0
Γ (2)
Γ (2 minus 120572)
1199091minus120572
Γ (3)
Γ (3 minus 120572)
1199092minus120572
sdot sdot sdot
Γ (119899 + 1)
Γ (119899 + 1 minus 120572)
119909119899minus120572
]A119879UΦ (119905)
=
[
[
[
[
[
[
[
[
[
[
[
1
119909
1199092
119909119899
]
]
]
]
]
]
]
]
]
]
]
119879
[
[
[
[
[
[
[
[
[
[
[
[
[
[
[
0 0 0 sdot sdot sdot 0
0
Γ (2)
Γ (2 minus 120572)
119909minus120572
0 sdot sdot sdot 0
0 0
Γ (3)
Γ (3 minus 120572)
119909minus120572
sdot sdot sdot 0
d 0
0 0 0 sdot sdot sdot
Γ (119899 + 1)
Γ (119899 + 1 minus 120572)
119909minus120572
]
]
]
]
]
]
]
]
]
]
]
]
]
]
]
A119879UΦ (119905) = Φ119879
(119909) (Aminus1)119879
MA119879UΦ (119905)
120597120573
119906 (119909 119905)
120597119905120573
≃
120597120573
(Φ119879
(119909)UΦ (119905))
120597119905120573
= Φ119879
(119909)U120597120573
Φ (119905)
120597119905120573
= Φ119879
(119909)UA120597120573T119899(119905)
120597119905120573
= Φ119879
(119909)UA [0
Γ (2)
Γ (2 minus 120573)
1199051minus120573
Γ (3)
Γ (3 minus 120573)
1199052minus120573
sdot sdot sdot
Γ (119899 + 1)
Γ (119899 + 1 minus 120573)
119905119899minus120573
]
119879
= Φ119879
(119909)UA
[
[
[
[
[
[
[
[
[
[
[
[
[
[
[
0 0 0 sdot sdot sdot 0
0
Γ (2)
Γ (2 minus 120573)
119905minus120573
0 sdot sdot sdot 0
0 0
Γ (3)
Γ (3 minus 120573)
119905minus120573
sdot sdot sdot 0
d 0
0 0 0 sdot sdot sdot
Γ (119899 + 1)
Γ (119899 + 1 minus 120573)
119905minus120573
]
]
]
]
]
]
]
]
]
]
]
]
]
]
]
T119899(119905) = Φ
119879
(119909)UANAminus1Φ (119905)
(19)
4 Mathematical Problems in Engineering
Let
M
=
[
[
[
[
[
[
[
[
[
[
[
[
[
[
0 0 0 sdot sdot sdot 0
0 Γ (2)Γ (2 minus 120572)
119909minus120572 0 sdot sdot sdot 0
0 0 Γ (3)Γ (3 minus 120572)
119909minus120572
sdot sdot sdot 0
d 0
0 0 0 sdot sdot sdot
Γ (119899 + 1)Γ (119899 + 1 minus 120572)
119909minus120572
]
]
]
]
]
]
]
]
]
]
]
]
]
]
N
=
[
[
[
[
[
[
[
[
[
[
[
[
[
[
[
0 0 0 sdot sdot sdot 0
0
Γ (2)
Γ (2 minus 120573)
119905minus120573
0 sdot sdot sdot 0
0 0
Γ (3)
Γ (3 minus 120573)
119905minus120573
sdot sdot sdot 0
d 0
0 0 0 sdot sdot sdot
Γ (119899 + 1)
Γ (119899 + 1 minus 120573)
119905minus120573
]
]
]
]
]
]
]
]
]
]
]
]
]
]
]
(20)
Substituting (19) into (1) we have
119886 (119909)Φ119879
(119909) (Aminus1)119879
MA119879UΦ (119905)
+ 119887 (119909)Φ119879
(119909)UANAminus1Φ (119905) = 119891 (119909 119905)
(21)
Dispersing (21) by the points (119909119894 119905119895) (119894 = 1 2 119899
119909 119895 =
1 2 119899119905) we can obtain U which is unknown
5 Error Analysis
In this part in order to illustrate the effectiveness of120597120572
119906(119909 119910)120597119909120572
cong Φ119879
(119909)UΦ(119910) we have given the followingtheorem Let 120597120572119906
119899(119909 119910)120597119909
120572 be the following approximationof 120597120572119906(119909 119910)120597119909
120572
120597120572
119906119899(119909 119910)
120597119909120572
=
119899
sum
119894=1
119899
sum
119895=1
119906119894119895119875119894(119909) 119875119895(119910) (22)
Then we have
120597120572
119906 (119909 119910)
120597119909120572
minus
120597120572
119906119899(119909 119910)
120597119909120572
=
infin
sum
119894=119899+1
infin
sum
119895=119899+1119906119894119895119875119894(119909) 119875119895(119910) (23)
Theorem4 Suppose that the function 120597120572
119906119899(119909 119910)120597119909
120572 obtain-ed by using Legendre polynomials is the approximation of120597120572
119906(119909 119910)120597119909120572 and 119906(119909 119910) has bounded mixed fractional
partial derivative |1205974+120572+120573
119906(119909 119910)1205971199092+120572
1205971199102+120573
| le then wehave the following upper bound of error
100381710038171003817100381710038171003817100381710038171003817
120597120572
119906 (119909 119910)
120597119909120572
minus
120597120572
119906119899(119909 119910)
120597119909120572
100381710038171003817100381710038171003817100381710038171003817119864
le
8
(
Γ1015840
(119899 minus 05)
Γ (119899 minus 05)
)
101584010158401015840
(24)
where 119906(119909 119910)119864= (int
1
minus1
int
1
minus1
1199062
(119909 119910)119889119909 119889119910)12 and
119906119894119895
= (
2119894 + 12
)(
2119895 + 12
)
sdot int
1
minus1int
1
minus1
120597120572
119906 (119909 119910)
120597119909120572
119875119894(119909) 119875119895(119910) 119889119909 119889119910
(25)
Proof Theproperty of the sequence 119875119894(119909) on [minus1 1] implies
that
int
1
minus1119875119894(119909) 119875119895(119909) 119889119909 =
22119894 + 1
119894 = 119895
0 119894 = 119895
(26)
then100381710038171003817100381710038171003817100381710038171003817
120597120572
119906 (119909 119910)
120597119909120572
minus
120597120572
119906119899(119909 119910)
120597119909120572
100381710038171003817100381710038171003817100381710038171003817
2
119864
= int
1
minus1int
1
minus1[
120597120572
119906 (119909 119910)
120597119909120572
minus
120597120572
119906119899(119909 119910)
120597119909120572
]
2
119889119909 119889119910
= int
1
minus1int
1
minus1[
[
infin
sum
119894=119899+1
infin
sum
119895=119899+1119906119894119895119875119894(119909) 119875119895(119910)
]
]
2
119889119909 119889119910
= int
1
minus1int
1
minus1
infin
sum
119894=119899+1
infin
sum
119895=119899+111990621198941198951198752119894(119909) 119875
2119895(119910) 119889119909 119889119910
=
infin
sum
119894=119899+1
infin
sum
119895=119899+11199062119894119895int
1
minus11198752119894(119909) 119889119909int
1
minus11198752119895(119910) 119889119910
=
infin
sum
119894=119899+1
infin
sum
119895=119899+11199062119894119895
22119894 + 1
22119895 + 1
(27)
The Legendre polynomials coefficients of function 120597120572
119906(119909 119910)
120597119909120572 are given by
119906119894119895
= (
2119894 + 12
)(
2119895 + 12
)
sdot int
1
minus1int
1
minus1
120597120572
119906 (119909 119910)
120597119909120572
119875119894(119909) 119875119895(119910) 119889119909 119889119910
(28)
Therefore we obtain
119906119894119895
=
2119895 + 14
int
1
minus1
120597120572
119906 (119909 119910)
120597119909120572
[119875119894+1 (119909) minus119875
119894minus1 (119909)]
sdot 119875119895(119910)
100381610038161003816100381610038161003816100381610038161003816
1
minus1119889119910minus
2119895 + 14
sdot int
1
minus1int
1
minus1
120597120572+1
119906 (119909 119910)
120597119909120572+1 [119875
119894+1 (119909) minus 119875119894minus1 (119909)] 119875119895 (119910) 119889119909 119889119910
= minus
2119895 + 14
int
1
minus1int
1
minus1
120597120572+1
119906 (119909 119910)
120597119909120572+1 [119875
119894+1 (119909)
minus 119875119894minus1 (119909)] 119875119895 (119910) 119889119909 119889119910 = minus
2119895 + 14
Mathematical Problems in Engineering 5
sdot int
1
minus1
120597120572+1
119906 (119909 119910)
120597119909120572+1 [
119875119894+2 (119909) minus 119875
119894(119909)
2119894 + 3
minus
119875119894(119909) minus 119875
119894minus2 (119909)
2119894 minus 1]119875119895(119910)
100381610038161003816100381610038161003816100381610038161003816
1
minus1119889119910+
2119895 + 14
sdot int
1
minus1int
1
minus1
120597120572+2
119906 (119909 119910)
120597119909120572+2 [
119875119894+2 (119909) minus 119875
119894(119909)
2119894 + 3
minus
119875119894(119909) minus 119875
119894minus2 (119909)
2119894 minus 1]119875119895(119910) 119889119909 119889119910 =
2119895 + 14
sdot int
1
minus1int
1
minus1
120597120572+2
119906 (119909 119910)
120597119909120572+2 [
119875119894+2 (119909) minus 119875
119894(119909)
2119894 + 3
minus
119875119894(119909) minus 119875
119894minus2 (119909)
2119894 minus 1]119875119895(119910) 119889119909 119889119910
(29)
Now let 120591119894(119909) = (2119894minus1)119875
119894+2(119909)minus2(2119894+1)119875119894(119909)+(2119894+3)119875119894minus2(119909)
then we have
119906119894119895
=
2119895 + 1
4 (2119894 minus 1) (2119894 + 3)
sdot int
1
minus1
int
1
minus1
1205972+120572
119906 (119909 119910)
1205971199092+120572
120591119894(119909) 119875119895(119910) 119889119909 119889119910
(30)
By solving this equation we have
119906119894119895
=
1
4 (2119894 minus 1) (2119894 + 3) (2119895 minus 1) (2119895 + 3)
sdot int
1
minus1
int
1
minus1
1205974+120572+120573
119906 (119909 119910)
1205971199092+120572
1205971199102+120573
120591119894(119909) 120591119895(119910) 119889119909 119889119910
(31)
So we have
10038161003816100381610038161003816119906119894119895
10038161003816100381610038161003816le
14 (2119894 minus 1) (2119894 + 3) (2119895 minus 1) (2119895 + 3)
sdot int
1
minus1int
1
minus1
1003816100381610038161003816100381610038161003816100381610038161003816
1205974+120572+120573
119906 (119909 119910)
1205971199092+120572
1205971199102+120573
1003816100381610038161003816100381610038161003816100381610038161003816
1003816100381610038161003816120591119894(119909)
1003816100381610038161003816
10038161003816100381610038161003816120591119895(119910)
10038161003816100381610038161003816119889119909 119889119910
le
4 (2119894 minus 1) (2119894 + 3) (2119895 minus 1) (2119895 + 3)
sdot int
1
minus1
1003816100381610038161003816120591119894(119909)
1003816100381610038161003816119889119909int
1
minus1
10038161003816100381610038161003816120591119895(119910)
10038161003816100381610038161003816119889119910
(32)
Moreover it was easily obtained that
int
1
minus1
1003816100381610038161003816120591119898
(119905)1003816100381610038161003816119889119905 le radic24 2119894 + 3
radic2119894 minus 3 (33)
thus we have
10038161003816100381610038161003816119906119894119895
10038161003816100381610038161003816le
244 (2119894 minus 1) (2119894 + 3) (2119895 minus 1) (2119895 + 3)
(2119894 + 3)radic2119894 minus 3
sdot
(2119895 + 3)radic2119895 minus 3
le
6(2119894 minus 3)32 (2119895 minus 3)32
(34)
Namely
10038161003816100381610038161003816119906119894119895
10038161003816100381610038161003816
2le
362
(2119894 minus 3)3 (2119895 minus 3)3 (35)
Therefore we have
100381710038171003817100381710038171003817100381710038171003817
120597120572
119906 (119909 119910)
120597119909120572
minus
120597120572
119906119899(119909 119910)
120597119909120572
100381710038171003817100381710038171003817100381710038171003817
2
119864
le
infin
sum
119894=119899+1
infin
sum
119895=119899+11199062119894119895
22119894 + 1
22119895 + 1
le
infin
sum
119894=119899+1
infin
sum
119895=119899+1
1442
(2119894 minus 3)3 (2119895 minus 3)3 (2119894 + 1) (2119895 + 1)
le
infin
sum
119894=119899+1
infin
sum
119895=119899+1
1442
(2119894 minus 3)4 (2119895 minus 3)4
= [
infin
sum
119894=119899+1
12(2119894 minus 3)4
]
2
= [
8(
Γ1015840
(119899 minus 05)Γ (119899 minus 05)
)
101584010158401015840
]
2
(36)
thus
100381710038171003817100381710038171003817100381710038171003817
120597120572
119906 (119909 119910)
120597119909120572
minus
120597120572
119906119899(119909 119910)
120597119909120572
100381710038171003817100381710038171003817100381710038171003817119864
le
8(
Γ1015840
(119899 minus 05)Γ (119899 minus 05)
)
101584010158401015840
(37)
This completes the proof
6 Numerical Examples
Example 1 Consider the following nonhomogeneous partialdifferential equation
11990913 120597
12119906 (119909 119905)
12059711990912 +119909
12 12059712
119906 (119909 119905)
12059711990512 = 119891 (119909 119905)
(119909 119905) isin [0 1] times [0 1] 119906 (119909 0) = 101199092
(1 minus 119909) 119906 (0 119905) = 119906 (1 119905) = 0
(38)
where 119891(119909 119905) = minus40radic119905(3 + 2119905)(minus1 + 119909)11990973
3radic120587 minus 16(1 +
119905)2
1199092
(minus5 + 6119909)3radic120587 The exact solution of this equation is119906(119909 119905) = 10119909
2
(1 minus 119909)(1 + 119905)2 Tables 1ndash3 show the absolute
errors for 119905 = 14119904 119905 = 12119904 and 119905 = 34119904 of different 119899
6 Mathematical Problems in Engineering
Table 1 Absolute error for t = 14 s and different values of 119899
119909 n = 2 n = 3 n = 401 03246 32831e minus 015 33582e minus 01602 02351 42734e minus 015 52745e minus 01503 04820 41237e minus 015 56521e minus 01504 03274 57320e minus 015 62742e minus 01605 08231 47381e minus 015 71640e minus 01606 09127 73722e minus 015 32356e minus 01607 12188 63276e minus 015 42745e minus 01508 15181 12374e minus 014 37224e minus 01509 08364 31744e minus 015 88874e minus 015
Table 2 Absolute error for t = 12 s and different values of 119899
119909 n = 2 n = 3 n = 401 02821 42137e minus 015 34325e minus 01602 04375 58711e minus 015 65332e minus 01603 01021 42210e minus 015 52435e minus 01604 02387 31016e minus 016 49722e minus 01605 08277 34762e minus 016 53478e minus 01506 09322 53265e minus 015 13241e minus 01607 13846 47632e minus 015 92371e minus 01608 16654 42346e minus 016 84812e minus 01509 09144 20437e minus 016 63273e minus 016
Table 3 Absolute error for t = 34 s and different values of 119899
119909 n = 2 n = 3 n = 401 03126 50832e minus 016 53281e minus 01602 08978 41845e minus 016 23258e minus 01503 02374 23448e minus 016 40112e minus 01604 02951 32155e minus 016 51223e minus 01605 03327 55518e minus 015 63274e minus 01506 13267 62440e minus 016 57421e minus 01507 08723 35220e minus 016 52871e minus 01608 09229 42301e minus 016 48810e minus 01609 11327 52310e minus 015 61138e minus 016
From Tables 1ndash3 we can see that the absolute error is verysmall when 119899 ge 3 Also when 119899 is fixed the more pointswe take the more accurate numerical solutions we obtainFigures 1ndash3 show the fact that 119899
119909119905is the number of 119909
119894 119905119895
Example 2 Consider the following fractional partial differ-ential equation
119909
12059714
119906 (119909 119905)
12059711990914 +119909
23 12059713
119906 (119909 119905)
12059711990513 = 119891 (119909 119905)
119906 (119909 0) = 119909 (119909 minus 1) (1199092
+ 1) 119906 (0 119905) = 0
(39)
0
2
4
6
002
0406
081
t0
0204
0608
1
x
u(xt)
Figure 1 Numerical solution of 119899119909119905
= 3
0
2
4
6
002
0406
081
t0
0204
0608
1
x
u(xt)
Figure 2 Numerical solution of 119899119909119905
= 6
0
05
1
00204
06081
0
2
4
6
t
x
u(xt)
Figure 3 Exact solution
where
119891 (119909 119905) =
311990523 (55 + 66119905 + 811199053) (minus1 + 119909) 1199092(1 + 119909
2)
110Γ (23)
+
4 (1 + 119905 + 1199052+ 119905
4) 119909
1712[minus385 + 8119909 (55 minus 60119909 + 641199092
)]
1155Γ (34)
(40)
Mathematical Problems in Engineering 7
0020406081 005
1
0
x t
minus02
minus04
minus06
minus08
minus1
minus12
minus14
u(xt)
Figure 4 The numerical solutions for Example 2 when 119899 = 4
00204060810
051
0
xt
minus02
minus04
minus06
minus08
minus1
minus12
minus14
u(xt)
Figure 5 The numerical solutions for Example 2 when 119899 = 5
The exact solution is 119906(119909 119905) = 119909(119909minus 1)(1199092
+1)(1 + 119905 + 1199052
+ 1199054
)The numerical solutions for 119899 = 4 119899 = 5 are displayed inFigures 4 and 5 and the exact solution is shown in Figure 6
Example 3 Consider this equation
11990912
12059713
119906 (119909 119905)
12059711990913
+11990923
12059713
119906 (119909 119905)
12059711990513
= 119891 (119909 119905)
119906 (119909 0) = 1199092
minus 1 119906 (0 119905) = minus1 minus (119905 minus 1)2
minus (119905 minus 1)3
(41)
where119891 (119909 119905)
=
9 (1 + 119905 minus 21199052 + 1199053) 119909
136
5Γ (23)
+
311990523 (minus2 + 3119905) (minus10 + 9119905) (minus1 + 1199092) 119909
23
40Γ (23)
(42)
The exact solution is 119906(119909 119905) = (1199092
minus 1)[1 + (119905 minus 1)2
+ (119905 minus 1)3
]The numerical solutions for 119899 = 3 119899 = 4 are displayed inFigures 7 and 8 The absolute errors are shown in Figures 9and 10 when 119899 = 3 and 119899 = 4
00204060810
05
1
0
xt
minus02
minus04
minus06
minus08
minus1
minus12
minus14
u(xt)
Figure 6 Exact solution for Example 2
0 02 04 06 08 1
0
05
1
0
u(xt)
x
t
minus02
minus04
minus06
minus08
minus1
minus12
minus14
Figure 7 The numerical solutions for Example 3 when 119899 = 3
0 02 04 06 08 1
0
05
1
0
minus02
minus04
minus06
minus08
minus1
minus12
minus14
u(xt)
xt
Figure 8 The numerical solutions for Example 3 when 119899 = 4
From Examples 1ndash3 we can see that the method in thispaper can be effectively used to solve the numerical solutionof fractional partial differential equation with variable coeffi-cients From the above results the absolute errors between
8 Mathematical Problems in Engineering
002
0406
081 0
05
10020406081
12
tx
times10minus14
u(xt)
Figure 9 The absolute errors for Example 3 when 119899 = 3
002
0406
081 0
0204
0608
10
020406081
12
tx
times10minus15
u(xt)
Figure 10 The absolute errors for Example 3 when 119899 = 4
the numerical solutions and the exact solution are rathersmall What is more due to the absolute error in this paperis about 10minus15 the Legendre polynomials method can reachhigher degree of accuracy by comparing the approximationsobtained by block pulse method [16]
7 Conclusion
In this paper we use the Legendre polynomials method tosolve a class of fractional partial differential equations withvariable coefficients The Legendre polynomials operationalmatrix of fractional differentiation is derived from the prop-erty of Legendre polynomials The initial equation is trans-lated into the product of some relevant matrixes which canalso be regarded as the system of linear equations The erroranalysis of Legendre polynomials is also given The numer-ical results show that numerical solutions obtained by ourmethod are in very good agreement with the exact solution
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
References
[1] R L Bagley and R A Calico ldquoFractional order state equationsfor the control of viscoelastically damped structuresrdquo Journalof Guidance Control and Dynamics vol 14 no 2 pp 304ndash3111991
[2] Z C Li and J S Luo Wavelet Analysis and Its ApplicationElectronic Industrial Publication Beijing China 2005
[3] J H Chen ldquoAnalysis of stability and convergence of numeri-cal approximation for the Riesz fractional reaction-dispersionequationrdquo Journal of Xiamen University (Natural Science) vol46 no 5 pp 616ndash619 2007
[4] D Delbosco and L Rodino ldquoExistence and uniqueness for anonlinear fractional differential equationrdquo Journal of Mathe-matical Analysis and Applications vol 204 no 2 pp 609ndash6251996
[5] Y Ryabov and A Puzenko ldquoA damped oscillations in view ofthe fraction oscillator equationrdquo Physics Review B vol 66 pp184ndash201 2002
[6] A M El-Sayed ldquoNonlinear functional-differential equationsof arbitrary ordersrdquo Nonlinear Analysis Theory Methods ampApplications vol 33 no 2 pp 181ndash186 1998
[7] I L El-Kalla ldquoError estimate of the series solution to a class ofnonlinear fractional differential equationsrdquo Communications inNonlinear Science and Numerical Simulation vol 16 no 3 pp1408ndash1413 2011
[8] Z M Odibat ldquoA study on the convergence of variationaliteration methodrdquo Mathematical and Computer Modelling vol51 no 9-10 pp 1181ndash1192 2010
[9] SMomani ZOdibat andV S Erturk ldquoGeneralized differentialtransform method for solving a space- and time-fractionaldiffusion-wave equationrdquo Physics Letters A vol 370 no 5-6 pp379ndash387 2007
[10] ZOdibat SMomani andV S Erturk ldquoGeneralized differentialtransform method application to differential equations offractional orderrdquo Applied Mathematics and Computation vol197 no 2 pp 467ndash477 2008
[11] Z Odibat and S Momani ldquoA generalized differential transformmethod for linear partial differential equations of fractionalorderrdquo Applied Mathematics Letters vol 21 no 2 pp 194ndash1992008
[12] Y Zhang ldquoA finite difference method for fractional partialdifferential equationrdquo Applied Mathematics and Computationvol 215 no 2 pp 524ndash529 2009
[13] Y X Wang and Q B Fan ldquoThe second kind Chebyshev waveletmethod for solving fractional differential equationsrdquo AppliedMathematics and Computation vol 218 no 17 pp 8592ndash86012012
[14] M X Yi and Y M Chen ldquoHaar wavelet operational matrixmethod for solving fractional partial differential equationsrdquoComputer Modeling in Engineering amp Sciences vol 88 no 3 pp229ndash244 2012
[15] EHDohaAH BhrawyD Baleanu and S S Ezz-Eldien ldquoTheoperational matrix formulation of the Jacobi tau approximationfor space fractional diffusion equationrdquo Advances in DifferenceEquations vol 231 pp 1687ndash1847 2014
[16] M X Yi J Huang and J X Wei ldquoBlock pulse operationalmatrix method for solving fractional partial differential equa-tionrdquo Applied Mathematics and Computation vol 221 pp 121ndash131 2013
Mathematical Problems in Engineering 9
[17] I Podlubny Fractional Differential Equations vol 198 ofMath-ematics in Science and Engineering Academic Press New YorkNY USA 1999
[18] A Saadatmandi and M Dehghan ldquoA new operational matrixfor solving fractional-order differential equationsrdquo ComputersandMathematics with Applications vol 59 no 3 pp 1326ndash13362010
[19] N Liu and E-B Lin ldquoLegendre wavelet method for numericalsolutions of partial differential equationsrdquo Numerical Methodsfor Partial Differential Equations vol 26 no 1 pp 81ndash94 2010
[20] S Nemati and Y Ordokhani ldquoLegendre expansion methodsfor the numerical solution of nonlinear 2D Fredholm integralequations of the second kindrdquo Journal of Applied Mathematicsamp Informatics vol 31 no 5-6 pp 609ndash621 2013
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
4 Mathematical Problems in Engineering
Let
M
=
[
[
[
[
[
[
[
[
[
[
[
[
[
[
0 0 0 sdot sdot sdot 0
0 Γ (2)Γ (2 minus 120572)
119909minus120572 0 sdot sdot sdot 0
0 0 Γ (3)Γ (3 minus 120572)
119909minus120572
sdot sdot sdot 0
d 0
0 0 0 sdot sdot sdot
Γ (119899 + 1)Γ (119899 + 1 minus 120572)
119909minus120572
]
]
]
]
]
]
]
]
]
]
]
]
]
]
N
=
[
[
[
[
[
[
[
[
[
[
[
[
[
[
[
0 0 0 sdot sdot sdot 0
0
Γ (2)
Γ (2 minus 120573)
119905minus120573
0 sdot sdot sdot 0
0 0
Γ (3)
Γ (3 minus 120573)
119905minus120573
sdot sdot sdot 0
d 0
0 0 0 sdot sdot sdot
Γ (119899 + 1)
Γ (119899 + 1 minus 120573)
119905minus120573
]
]
]
]
]
]
]
]
]
]
]
]
]
]
]
(20)
Substituting (19) into (1) we have
119886 (119909)Φ119879
(119909) (Aminus1)119879
MA119879UΦ (119905)
+ 119887 (119909)Φ119879
(119909)UANAminus1Φ (119905) = 119891 (119909 119905)
(21)
Dispersing (21) by the points (119909119894 119905119895) (119894 = 1 2 119899
119909 119895 =
1 2 119899119905) we can obtain U which is unknown
5 Error Analysis
In this part in order to illustrate the effectiveness of120597120572
119906(119909 119910)120597119909120572
cong Φ119879
(119909)UΦ(119910) we have given the followingtheorem Let 120597120572119906
119899(119909 119910)120597119909
120572 be the following approximationof 120597120572119906(119909 119910)120597119909
120572
120597120572
119906119899(119909 119910)
120597119909120572
=
119899
sum
119894=1
119899
sum
119895=1
119906119894119895119875119894(119909) 119875119895(119910) (22)
Then we have
120597120572
119906 (119909 119910)
120597119909120572
minus
120597120572
119906119899(119909 119910)
120597119909120572
=
infin
sum
119894=119899+1
infin
sum
119895=119899+1119906119894119895119875119894(119909) 119875119895(119910) (23)
Theorem4 Suppose that the function 120597120572
119906119899(119909 119910)120597119909
120572 obtain-ed by using Legendre polynomials is the approximation of120597120572
119906(119909 119910)120597119909120572 and 119906(119909 119910) has bounded mixed fractional
partial derivative |1205974+120572+120573
119906(119909 119910)1205971199092+120572
1205971199102+120573
| le then wehave the following upper bound of error
100381710038171003817100381710038171003817100381710038171003817
120597120572
119906 (119909 119910)
120597119909120572
minus
120597120572
119906119899(119909 119910)
120597119909120572
100381710038171003817100381710038171003817100381710038171003817119864
le
8
(
Γ1015840
(119899 minus 05)
Γ (119899 minus 05)
)
101584010158401015840
(24)
where 119906(119909 119910)119864= (int
1
minus1
int
1
minus1
1199062
(119909 119910)119889119909 119889119910)12 and
119906119894119895
= (
2119894 + 12
)(
2119895 + 12
)
sdot int
1
minus1int
1
minus1
120597120572
119906 (119909 119910)
120597119909120572
119875119894(119909) 119875119895(119910) 119889119909 119889119910
(25)
Proof Theproperty of the sequence 119875119894(119909) on [minus1 1] implies
that
int
1
minus1119875119894(119909) 119875119895(119909) 119889119909 =
22119894 + 1
119894 = 119895
0 119894 = 119895
(26)
then100381710038171003817100381710038171003817100381710038171003817
120597120572
119906 (119909 119910)
120597119909120572
minus
120597120572
119906119899(119909 119910)
120597119909120572
100381710038171003817100381710038171003817100381710038171003817
2
119864
= int
1
minus1int
1
minus1[
120597120572
119906 (119909 119910)
120597119909120572
minus
120597120572
119906119899(119909 119910)
120597119909120572
]
2
119889119909 119889119910
= int
1
minus1int
1
minus1[
[
infin
sum
119894=119899+1
infin
sum
119895=119899+1119906119894119895119875119894(119909) 119875119895(119910)
]
]
2
119889119909 119889119910
= int
1
minus1int
1
minus1
infin
sum
119894=119899+1
infin
sum
119895=119899+111990621198941198951198752119894(119909) 119875
2119895(119910) 119889119909 119889119910
=
infin
sum
119894=119899+1
infin
sum
119895=119899+11199062119894119895int
1
minus11198752119894(119909) 119889119909int
1
minus11198752119895(119910) 119889119910
=
infin
sum
119894=119899+1
infin
sum
119895=119899+11199062119894119895
22119894 + 1
22119895 + 1
(27)
The Legendre polynomials coefficients of function 120597120572
119906(119909 119910)
120597119909120572 are given by
119906119894119895
= (
2119894 + 12
)(
2119895 + 12
)
sdot int
1
minus1int
1
minus1
120597120572
119906 (119909 119910)
120597119909120572
119875119894(119909) 119875119895(119910) 119889119909 119889119910
(28)
Therefore we obtain
119906119894119895
=
2119895 + 14
int
1
minus1
120597120572
119906 (119909 119910)
120597119909120572
[119875119894+1 (119909) minus119875
119894minus1 (119909)]
sdot 119875119895(119910)
100381610038161003816100381610038161003816100381610038161003816
1
minus1119889119910minus
2119895 + 14
sdot int
1
minus1int
1
minus1
120597120572+1
119906 (119909 119910)
120597119909120572+1 [119875
119894+1 (119909) minus 119875119894minus1 (119909)] 119875119895 (119910) 119889119909 119889119910
= minus
2119895 + 14
int
1
minus1int
1
minus1
120597120572+1
119906 (119909 119910)
120597119909120572+1 [119875
119894+1 (119909)
minus 119875119894minus1 (119909)] 119875119895 (119910) 119889119909 119889119910 = minus
2119895 + 14
Mathematical Problems in Engineering 5
sdot int
1
minus1
120597120572+1
119906 (119909 119910)
120597119909120572+1 [
119875119894+2 (119909) minus 119875
119894(119909)
2119894 + 3
minus
119875119894(119909) minus 119875
119894minus2 (119909)
2119894 minus 1]119875119895(119910)
100381610038161003816100381610038161003816100381610038161003816
1
minus1119889119910+
2119895 + 14
sdot int
1
minus1int
1
minus1
120597120572+2
119906 (119909 119910)
120597119909120572+2 [
119875119894+2 (119909) minus 119875
119894(119909)
2119894 + 3
minus
119875119894(119909) minus 119875
119894minus2 (119909)
2119894 minus 1]119875119895(119910) 119889119909 119889119910 =
2119895 + 14
sdot int
1
minus1int
1
minus1
120597120572+2
119906 (119909 119910)
120597119909120572+2 [
119875119894+2 (119909) minus 119875
119894(119909)
2119894 + 3
minus
119875119894(119909) minus 119875
119894minus2 (119909)
2119894 minus 1]119875119895(119910) 119889119909 119889119910
(29)
Now let 120591119894(119909) = (2119894minus1)119875
119894+2(119909)minus2(2119894+1)119875119894(119909)+(2119894+3)119875119894minus2(119909)
then we have
119906119894119895
=
2119895 + 1
4 (2119894 minus 1) (2119894 + 3)
sdot int
1
minus1
int
1
minus1
1205972+120572
119906 (119909 119910)
1205971199092+120572
120591119894(119909) 119875119895(119910) 119889119909 119889119910
(30)
By solving this equation we have
119906119894119895
=
1
4 (2119894 minus 1) (2119894 + 3) (2119895 minus 1) (2119895 + 3)
sdot int
1
minus1
int
1
minus1
1205974+120572+120573
119906 (119909 119910)
1205971199092+120572
1205971199102+120573
120591119894(119909) 120591119895(119910) 119889119909 119889119910
(31)
So we have
10038161003816100381610038161003816119906119894119895
10038161003816100381610038161003816le
14 (2119894 minus 1) (2119894 + 3) (2119895 minus 1) (2119895 + 3)
sdot int
1
minus1int
1
minus1
1003816100381610038161003816100381610038161003816100381610038161003816
1205974+120572+120573
119906 (119909 119910)
1205971199092+120572
1205971199102+120573
1003816100381610038161003816100381610038161003816100381610038161003816
1003816100381610038161003816120591119894(119909)
1003816100381610038161003816
10038161003816100381610038161003816120591119895(119910)
10038161003816100381610038161003816119889119909 119889119910
le
4 (2119894 minus 1) (2119894 + 3) (2119895 minus 1) (2119895 + 3)
sdot int
1
minus1
1003816100381610038161003816120591119894(119909)
1003816100381610038161003816119889119909int
1
minus1
10038161003816100381610038161003816120591119895(119910)
10038161003816100381610038161003816119889119910
(32)
Moreover it was easily obtained that
int
1
minus1
1003816100381610038161003816120591119898
(119905)1003816100381610038161003816119889119905 le radic24 2119894 + 3
radic2119894 minus 3 (33)
thus we have
10038161003816100381610038161003816119906119894119895
10038161003816100381610038161003816le
244 (2119894 minus 1) (2119894 + 3) (2119895 minus 1) (2119895 + 3)
(2119894 + 3)radic2119894 minus 3
sdot
(2119895 + 3)radic2119895 minus 3
le
6(2119894 minus 3)32 (2119895 minus 3)32
(34)
Namely
10038161003816100381610038161003816119906119894119895
10038161003816100381610038161003816
2le
362
(2119894 minus 3)3 (2119895 minus 3)3 (35)
Therefore we have
100381710038171003817100381710038171003817100381710038171003817
120597120572
119906 (119909 119910)
120597119909120572
minus
120597120572
119906119899(119909 119910)
120597119909120572
100381710038171003817100381710038171003817100381710038171003817
2
119864
le
infin
sum
119894=119899+1
infin
sum
119895=119899+11199062119894119895
22119894 + 1
22119895 + 1
le
infin
sum
119894=119899+1
infin
sum
119895=119899+1
1442
(2119894 minus 3)3 (2119895 minus 3)3 (2119894 + 1) (2119895 + 1)
le
infin
sum
119894=119899+1
infin
sum
119895=119899+1
1442
(2119894 minus 3)4 (2119895 minus 3)4
= [
infin
sum
119894=119899+1
12(2119894 minus 3)4
]
2
= [
8(
Γ1015840
(119899 minus 05)Γ (119899 minus 05)
)
101584010158401015840
]
2
(36)
thus
100381710038171003817100381710038171003817100381710038171003817
120597120572
119906 (119909 119910)
120597119909120572
minus
120597120572
119906119899(119909 119910)
120597119909120572
100381710038171003817100381710038171003817100381710038171003817119864
le
8(
Γ1015840
(119899 minus 05)Γ (119899 minus 05)
)
101584010158401015840
(37)
This completes the proof
6 Numerical Examples
Example 1 Consider the following nonhomogeneous partialdifferential equation
11990913 120597
12119906 (119909 119905)
12059711990912 +119909
12 12059712
119906 (119909 119905)
12059711990512 = 119891 (119909 119905)
(119909 119905) isin [0 1] times [0 1] 119906 (119909 0) = 101199092
(1 minus 119909) 119906 (0 119905) = 119906 (1 119905) = 0
(38)
where 119891(119909 119905) = minus40radic119905(3 + 2119905)(minus1 + 119909)11990973
3radic120587 minus 16(1 +
119905)2
1199092
(minus5 + 6119909)3radic120587 The exact solution of this equation is119906(119909 119905) = 10119909
2
(1 minus 119909)(1 + 119905)2 Tables 1ndash3 show the absolute
errors for 119905 = 14119904 119905 = 12119904 and 119905 = 34119904 of different 119899
6 Mathematical Problems in Engineering
Table 1 Absolute error for t = 14 s and different values of 119899
119909 n = 2 n = 3 n = 401 03246 32831e minus 015 33582e minus 01602 02351 42734e minus 015 52745e minus 01503 04820 41237e minus 015 56521e minus 01504 03274 57320e minus 015 62742e minus 01605 08231 47381e minus 015 71640e minus 01606 09127 73722e minus 015 32356e minus 01607 12188 63276e minus 015 42745e minus 01508 15181 12374e minus 014 37224e minus 01509 08364 31744e minus 015 88874e minus 015
Table 2 Absolute error for t = 12 s and different values of 119899
119909 n = 2 n = 3 n = 401 02821 42137e minus 015 34325e minus 01602 04375 58711e minus 015 65332e minus 01603 01021 42210e minus 015 52435e minus 01604 02387 31016e minus 016 49722e minus 01605 08277 34762e minus 016 53478e minus 01506 09322 53265e minus 015 13241e minus 01607 13846 47632e minus 015 92371e minus 01608 16654 42346e minus 016 84812e minus 01509 09144 20437e minus 016 63273e minus 016
Table 3 Absolute error for t = 34 s and different values of 119899
119909 n = 2 n = 3 n = 401 03126 50832e minus 016 53281e minus 01602 08978 41845e minus 016 23258e minus 01503 02374 23448e minus 016 40112e minus 01604 02951 32155e minus 016 51223e minus 01605 03327 55518e minus 015 63274e minus 01506 13267 62440e minus 016 57421e minus 01507 08723 35220e minus 016 52871e minus 01608 09229 42301e minus 016 48810e minus 01609 11327 52310e minus 015 61138e minus 016
From Tables 1ndash3 we can see that the absolute error is verysmall when 119899 ge 3 Also when 119899 is fixed the more pointswe take the more accurate numerical solutions we obtainFigures 1ndash3 show the fact that 119899
119909119905is the number of 119909
119894 119905119895
Example 2 Consider the following fractional partial differ-ential equation
119909
12059714
119906 (119909 119905)
12059711990914 +119909
23 12059713
119906 (119909 119905)
12059711990513 = 119891 (119909 119905)
119906 (119909 0) = 119909 (119909 minus 1) (1199092
+ 1) 119906 (0 119905) = 0
(39)
0
2
4
6
002
0406
081
t0
0204
0608
1
x
u(xt)
Figure 1 Numerical solution of 119899119909119905
= 3
0
2
4
6
002
0406
081
t0
0204
0608
1
x
u(xt)
Figure 2 Numerical solution of 119899119909119905
= 6
0
05
1
00204
06081
0
2
4
6
t
x
u(xt)
Figure 3 Exact solution
where
119891 (119909 119905) =
311990523 (55 + 66119905 + 811199053) (minus1 + 119909) 1199092(1 + 119909
2)
110Γ (23)
+
4 (1 + 119905 + 1199052+ 119905
4) 119909
1712[minus385 + 8119909 (55 minus 60119909 + 641199092
)]
1155Γ (34)
(40)
Mathematical Problems in Engineering 7
0020406081 005
1
0
x t
minus02
minus04
minus06
minus08
minus1
minus12
minus14
u(xt)
Figure 4 The numerical solutions for Example 2 when 119899 = 4
00204060810
051
0
xt
minus02
minus04
minus06
minus08
minus1
minus12
minus14
u(xt)
Figure 5 The numerical solutions for Example 2 when 119899 = 5
The exact solution is 119906(119909 119905) = 119909(119909minus 1)(1199092
+1)(1 + 119905 + 1199052
+ 1199054
)The numerical solutions for 119899 = 4 119899 = 5 are displayed inFigures 4 and 5 and the exact solution is shown in Figure 6
Example 3 Consider this equation
11990912
12059713
119906 (119909 119905)
12059711990913
+11990923
12059713
119906 (119909 119905)
12059711990513
= 119891 (119909 119905)
119906 (119909 0) = 1199092
minus 1 119906 (0 119905) = minus1 minus (119905 minus 1)2
minus (119905 minus 1)3
(41)
where119891 (119909 119905)
=
9 (1 + 119905 minus 21199052 + 1199053) 119909
136
5Γ (23)
+
311990523 (minus2 + 3119905) (minus10 + 9119905) (minus1 + 1199092) 119909
23
40Γ (23)
(42)
The exact solution is 119906(119909 119905) = (1199092
minus 1)[1 + (119905 minus 1)2
+ (119905 minus 1)3
]The numerical solutions for 119899 = 3 119899 = 4 are displayed inFigures 7 and 8 The absolute errors are shown in Figures 9and 10 when 119899 = 3 and 119899 = 4
00204060810
05
1
0
xt
minus02
minus04
minus06
minus08
minus1
minus12
minus14
u(xt)
Figure 6 Exact solution for Example 2
0 02 04 06 08 1
0
05
1
0
u(xt)
x
t
minus02
minus04
minus06
minus08
minus1
minus12
minus14
Figure 7 The numerical solutions for Example 3 when 119899 = 3
0 02 04 06 08 1
0
05
1
0
minus02
minus04
minus06
minus08
minus1
minus12
minus14
u(xt)
xt
Figure 8 The numerical solutions for Example 3 when 119899 = 4
From Examples 1ndash3 we can see that the method in thispaper can be effectively used to solve the numerical solutionof fractional partial differential equation with variable coeffi-cients From the above results the absolute errors between
8 Mathematical Problems in Engineering
002
0406
081 0
05
10020406081
12
tx
times10minus14
u(xt)
Figure 9 The absolute errors for Example 3 when 119899 = 3
002
0406
081 0
0204
0608
10
020406081
12
tx
times10minus15
u(xt)
Figure 10 The absolute errors for Example 3 when 119899 = 4
the numerical solutions and the exact solution are rathersmall What is more due to the absolute error in this paperis about 10minus15 the Legendre polynomials method can reachhigher degree of accuracy by comparing the approximationsobtained by block pulse method [16]
7 Conclusion
In this paper we use the Legendre polynomials method tosolve a class of fractional partial differential equations withvariable coefficients The Legendre polynomials operationalmatrix of fractional differentiation is derived from the prop-erty of Legendre polynomials The initial equation is trans-lated into the product of some relevant matrixes which canalso be regarded as the system of linear equations The erroranalysis of Legendre polynomials is also given The numer-ical results show that numerical solutions obtained by ourmethod are in very good agreement with the exact solution
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
References
[1] R L Bagley and R A Calico ldquoFractional order state equationsfor the control of viscoelastically damped structuresrdquo Journalof Guidance Control and Dynamics vol 14 no 2 pp 304ndash3111991
[2] Z C Li and J S Luo Wavelet Analysis and Its ApplicationElectronic Industrial Publication Beijing China 2005
[3] J H Chen ldquoAnalysis of stability and convergence of numeri-cal approximation for the Riesz fractional reaction-dispersionequationrdquo Journal of Xiamen University (Natural Science) vol46 no 5 pp 616ndash619 2007
[4] D Delbosco and L Rodino ldquoExistence and uniqueness for anonlinear fractional differential equationrdquo Journal of Mathe-matical Analysis and Applications vol 204 no 2 pp 609ndash6251996
[5] Y Ryabov and A Puzenko ldquoA damped oscillations in view ofthe fraction oscillator equationrdquo Physics Review B vol 66 pp184ndash201 2002
[6] A M El-Sayed ldquoNonlinear functional-differential equationsof arbitrary ordersrdquo Nonlinear Analysis Theory Methods ampApplications vol 33 no 2 pp 181ndash186 1998
[7] I L El-Kalla ldquoError estimate of the series solution to a class ofnonlinear fractional differential equationsrdquo Communications inNonlinear Science and Numerical Simulation vol 16 no 3 pp1408ndash1413 2011
[8] Z M Odibat ldquoA study on the convergence of variationaliteration methodrdquo Mathematical and Computer Modelling vol51 no 9-10 pp 1181ndash1192 2010
[9] SMomani ZOdibat andV S Erturk ldquoGeneralized differentialtransform method for solving a space- and time-fractionaldiffusion-wave equationrdquo Physics Letters A vol 370 no 5-6 pp379ndash387 2007
[10] ZOdibat SMomani andV S Erturk ldquoGeneralized differentialtransform method application to differential equations offractional orderrdquo Applied Mathematics and Computation vol197 no 2 pp 467ndash477 2008
[11] Z Odibat and S Momani ldquoA generalized differential transformmethod for linear partial differential equations of fractionalorderrdquo Applied Mathematics Letters vol 21 no 2 pp 194ndash1992008
[12] Y Zhang ldquoA finite difference method for fractional partialdifferential equationrdquo Applied Mathematics and Computationvol 215 no 2 pp 524ndash529 2009
[13] Y X Wang and Q B Fan ldquoThe second kind Chebyshev waveletmethod for solving fractional differential equationsrdquo AppliedMathematics and Computation vol 218 no 17 pp 8592ndash86012012
[14] M X Yi and Y M Chen ldquoHaar wavelet operational matrixmethod for solving fractional partial differential equationsrdquoComputer Modeling in Engineering amp Sciences vol 88 no 3 pp229ndash244 2012
[15] EHDohaAH BhrawyD Baleanu and S S Ezz-Eldien ldquoTheoperational matrix formulation of the Jacobi tau approximationfor space fractional diffusion equationrdquo Advances in DifferenceEquations vol 231 pp 1687ndash1847 2014
[16] M X Yi J Huang and J X Wei ldquoBlock pulse operationalmatrix method for solving fractional partial differential equa-tionrdquo Applied Mathematics and Computation vol 221 pp 121ndash131 2013
Mathematical Problems in Engineering 9
[17] I Podlubny Fractional Differential Equations vol 198 ofMath-ematics in Science and Engineering Academic Press New YorkNY USA 1999
[18] A Saadatmandi and M Dehghan ldquoA new operational matrixfor solving fractional-order differential equationsrdquo ComputersandMathematics with Applications vol 59 no 3 pp 1326ndash13362010
[19] N Liu and E-B Lin ldquoLegendre wavelet method for numericalsolutions of partial differential equationsrdquo Numerical Methodsfor Partial Differential Equations vol 26 no 1 pp 81ndash94 2010
[20] S Nemati and Y Ordokhani ldquoLegendre expansion methodsfor the numerical solution of nonlinear 2D Fredholm integralequations of the second kindrdquo Journal of Applied Mathematicsamp Informatics vol 31 no 5-6 pp 609ndash621 2013
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
Mathematical Problems in Engineering 5
sdot int
1
minus1
120597120572+1
119906 (119909 119910)
120597119909120572+1 [
119875119894+2 (119909) minus 119875
119894(119909)
2119894 + 3
minus
119875119894(119909) minus 119875
119894minus2 (119909)
2119894 minus 1]119875119895(119910)
100381610038161003816100381610038161003816100381610038161003816
1
minus1119889119910+
2119895 + 14
sdot int
1
minus1int
1
minus1
120597120572+2
119906 (119909 119910)
120597119909120572+2 [
119875119894+2 (119909) minus 119875
119894(119909)
2119894 + 3
minus
119875119894(119909) minus 119875
119894minus2 (119909)
2119894 minus 1]119875119895(119910) 119889119909 119889119910 =
2119895 + 14
sdot int
1
minus1int
1
minus1
120597120572+2
119906 (119909 119910)
120597119909120572+2 [
119875119894+2 (119909) minus 119875
119894(119909)
2119894 + 3
minus
119875119894(119909) minus 119875
119894minus2 (119909)
2119894 minus 1]119875119895(119910) 119889119909 119889119910
(29)
Now let 120591119894(119909) = (2119894minus1)119875
119894+2(119909)minus2(2119894+1)119875119894(119909)+(2119894+3)119875119894minus2(119909)
then we have
119906119894119895
=
2119895 + 1
4 (2119894 minus 1) (2119894 + 3)
sdot int
1
minus1
int
1
minus1
1205972+120572
119906 (119909 119910)
1205971199092+120572
120591119894(119909) 119875119895(119910) 119889119909 119889119910
(30)
By solving this equation we have
119906119894119895
=
1
4 (2119894 minus 1) (2119894 + 3) (2119895 minus 1) (2119895 + 3)
sdot int
1
minus1
int
1
minus1
1205974+120572+120573
119906 (119909 119910)
1205971199092+120572
1205971199102+120573
120591119894(119909) 120591119895(119910) 119889119909 119889119910
(31)
So we have
10038161003816100381610038161003816119906119894119895
10038161003816100381610038161003816le
14 (2119894 minus 1) (2119894 + 3) (2119895 minus 1) (2119895 + 3)
sdot int
1
minus1int
1
minus1
1003816100381610038161003816100381610038161003816100381610038161003816
1205974+120572+120573
119906 (119909 119910)
1205971199092+120572
1205971199102+120573
1003816100381610038161003816100381610038161003816100381610038161003816
1003816100381610038161003816120591119894(119909)
1003816100381610038161003816
10038161003816100381610038161003816120591119895(119910)
10038161003816100381610038161003816119889119909 119889119910
le
4 (2119894 minus 1) (2119894 + 3) (2119895 minus 1) (2119895 + 3)
sdot int
1
minus1
1003816100381610038161003816120591119894(119909)
1003816100381610038161003816119889119909int
1
minus1
10038161003816100381610038161003816120591119895(119910)
10038161003816100381610038161003816119889119910
(32)
Moreover it was easily obtained that
int
1
minus1
1003816100381610038161003816120591119898
(119905)1003816100381610038161003816119889119905 le radic24 2119894 + 3
radic2119894 minus 3 (33)
thus we have
10038161003816100381610038161003816119906119894119895
10038161003816100381610038161003816le
244 (2119894 minus 1) (2119894 + 3) (2119895 minus 1) (2119895 + 3)
(2119894 + 3)radic2119894 minus 3
sdot
(2119895 + 3)radic2119895 minus 3
le
6(2119894 minus 3)32 (2119895 minus 3)32
(34)
Namely
10038161003816100381610038161003816119906119894119895
10038161003816100381610038161003816
2le
362
(2119894 minus 3)3 (2119895 minus 3)3 (35)
Therefore we have
100381710038171003817100381710038171003817100381710038171003817
120597120572
119906 (119909 119910)
120597119909120572
minus
120597120572
119906119899(119909 119910)
120597119909120572
100381710038171003817100381710038171003817100381710038171003817
2
119864
le
infin
sum
119894=119899+1
infin
sum
119895=119899+11199062119894119895
22119894 + 1
22119895 + 1
le
infin
sum
119894=119899+1
infin
sum
119895=119899+1
1442
(2119894 minus 3)3 (2119895 minus 3)3 (2119894 + 1) (2119895 + 1)
le
infin
sum
119894=119899+1
infin
sum
119895=119899+1
1442
(2119894 minus 3)4 (2119895 minus 3)4
= [
infin
sum
119894=119899+1
12(2119894 minus 3)4
]
2
= [
8(
Γ1015840
(119899 minus 05)Γ (119899 minus 05)
)
101584010158401015840
]
2
(36)
thus
100381710038171003817100381710038171003817100381710038171003817
120597120572
119906 (119909 119910)
120597119909120572
minus
120597120572
119906119899(119909 119910)
120597119909120572
100381710038171003817100381710038171003817100381710038171003817119864
le
8(
Γ1015840
(119899 minus 05)Γ (119899 minus 05)
)
101584010158401015840
(37)
This completes the proof
6 Numerical Examples
Example 1 Consider the following nonhomogeneous partialdifferential equation
11990913 120597
12119906 (119909 119905)
12059711990912 +119909
12 12059712
119906 (119909 119905)
12059711990512 = 119891 (119909 119905)
(119909 119905) isin [0 1] times [0 1] 119906 (119909 0) = 101199092
(1 minus 119909) 119906 (0 119905) = 119906 (1 119905) = 0
(38)
where 119891(119909 119905) = minus40radic119905(3 + 2119905)(minus1 + 119909)11990973
3radic120587 minus 16(1 +
119905)2
1199092
(minus5 + 6119909)3radic120587 The exact solution of this equation is119906(119909 119905) = 10119909
2
(1 minus 119909)(1 + 119905)2 Tables 1ndash3 show the absolute
errors for 119905 = 14119904 119905 = 12119904 and 119905 = 34119904 of different 119899
6 Mathematical Problems in Engineering
Table 1 Absolute error for t = 14 s and different values of 119899
119909 n = 2 n = 3 n = 401 03246 32831e minus 015 33582e minus 01602 02351 42734e minus 015 52745e minus 01503 04820 41237e minus 015 56521e minus 01504 03274 57320e minus 015 62742e minus 01605 08231 47381e minus 015 71640e minus 01606 09127 73722e minus 015 32356e minus 01607 12188 63276e minus 015 42745e minus 01508 15181 12374e minus 014 37224e minus 01509 08364 31744e minus 015 88874e minus 015
Table 2 Absolute error for t = 12 s and different values of 119899
119909 n = 2 n = 3 n = 401 02821 42137e minus 015 34325e minus 01602 04375 58711e minus 015 65332e minus 01603 01021 42210e minus 015 52435e minus 01604 02387 31016e minus 016 49722e minus 01605 08277 34762e minus 016 53478e minus 01506 09322 53265e minus 015 13241e minus 01607 13846 47632e minus 015 92371e minus 01608 16654 42346e minus 016 84812e minus 01509 09144 20437e minus 016 63273e minus 016
Table 3 Absolute error for t = 34 s and different values of 119899
119909 n = 2 n = 3 n = 401 03126 50832e minus 016 53281e minus 01602 08978 41845e minus 016 23258e minus 01503 02374 23448e minus 016 40112e minus 01604 02951 32155e minus 016 51223e minus 01605 03327 55518e minus 015 63274e minus 01506 13267 62440e minus 016 57421e minus 01507 08723 35220e minus 016 52871e minus 01608 09229 42301e minus 016 48810e minus 01609 11327 52310e minus 015 61138e minus 016
From Tables 1ndash3 we can see that the absolute error is verysmall when 119899 ge 3 Also when 119899 is fixed the more pointswe take the more accurate numerical solutions we obtainFigures 1ndash3 show the fact that 119899
119909119905is the number of 119909
119894 119905119895
Example 2 Consider the following fractional partial differ-ential equation
119909
12059714
119906 (119909 119905)
12059711990914 +119909
23 12059713
119906 (119909 119905)
12059711990513 = 119891 (119909 119905)
119906 (119909 0) = 119909 (119909 minus 1) (1199092
+ 1) 119906 (0 119905) = 0
(39)
0
2
4
6
002
0406
081
t0
0204
0608
1
x
u(xt)
Figure 1 Numerical solution of 119899119909119905
= 3
0
2
4
6
002
0406
081
t0
0204
0608
1
x
u(xt)
Figure 2 Numerical solution of 119899119909119905
= 6
0
05
1
00204
06081
0
2
4
6
t
x
u(xt)
Figure 3 Exact solution
where
119891 (119909 119905) =
311990523 (55 + 66119905 + 811199053) (minus1 + 119909) 1199092(1 + 119909
2)
110Γ (23)
+
4 (1 + 119905 + 1199052+ 119905
4) 119909
1712[minus385 + 8119909 (55 minus 60119909 + 641199092
)]
1155Γ (34)
(40)
Mathematical Problems in Engineering 7
0020406081 005
1
0
x t
minus02
minus04
minus06
minus08
minus1
minus12
minus14
u(xt)
Figure 4 The numerical solutions for Example 2 when 119899 = 4
00204060810
051
0
xt
minus02
minus04
minus06
minus08
minus1
minus12
minus14
u(xt)
Figure 5 The numerical solutions for Example 2 when 119899 = 5
The exact solution is 119906(119909 119905) = 119909(119909minus 1)(1199092
+1)(1 + 119905 + 1199052
+ 1199054
)The numerical solutions for 119899 = 4 119899 = 5 are displayed inFigures 4 and 5 and the exact solution is shown in Figure 6
Example 3 Consider this equation
11990912
12059713
119906 (119909 119905)
12059711990913
+11990923
12059713
119906 (119909 119905)
12059711990513
= 119891 (119909 119905)
119906 (119909 0) = 1199092
minus 1 119906 (0 119905) = minus1 minus (119905 minus 1)2
minus (119905 minus 1)3
(41)
where119891 (119909 119905)
=
9 (1 + 119905 minus 21199052 + 1199053) 119909
136
5Γ (23)
+
311990523 (minus2 + 3119905) (minus10 + 9119905) (minus1 + 1199092) 119909
23
40Γ (23)
(42)
The exact solution is 119906(119909 119905) = (1199092
minus 1)[1 + (119905 minus 1)2
+ (119905 minus 1)3
]The numerical solutions for 119899 = 3 119899 = 4 are displayed inFigures 7 and 8 The absolute errors are shown in Figures 9and 10 when 119899 = 3 and 119899 = 4
00204060810
05
1
0
xt
minus02
minus04
minus06
minus08
minus1
minus12
minus14
u(xt)
Figure 6 Exact solution for Example 2
0 02 04 06 08 1
0
05
1
0
u(xt)
x
t
minus02
minus04
minus06
minus08
minus1
minus12
minus14
Figure 7 The numerical solutions for Example 3 when 119899 = 3
0 02 04 06 08 1
0
05
1
0
minus02
minus04
minus06
minus08
minus1
minus12
minus14
u(xt)
xt
Figure 8 The numerical solutions for Example 3 when 119899 = 4
From Examples 1ndash3 we can see that the method in thispaper can be effectively used to solve the numerical solutionof fractional partial differential equation with variable coeffi-cients From the above results the absolute errors between
8 Mathematical Problems in Engineering
002
0406
081 0
05
10020406081
12
tx
times10minus14
u(xt)
Figure 9 The absolute errors for Example 3 when 119899 = 3
002
0406
081 0
0204
0608
10
020406081
12
tx
times10minus15
u(xt)
Figure 10 The absolute errors for Example 3 when 119899 = 4
the numerical solutions and the exact solution are rathersmall What is more due to the absolute error in this paperis about 10minus15 the Legendre polynomials method can reachhigher degree of accuracy by comparing the approximationsobtained by block pulse method [16]
7 Conclusion
In this paper we use the Legendre polynomials method tosolve a class of fractional partial differential equations withvariable coefficients The Legendre polynomials operationalmatrix of fractional differentiation is derived from the prop-erty of Legendre polynomials The initial equation is trans-lated into the product of some relevant matrixes which canalso be regarded as the system of linear equations The erroranalysis of Legendre polynomials is also given The numer-ical results show that numerical solutions obtained by ourmethod are in very good agreement with the exact solution
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
References
[1] R L Bagley and R A Calico ldquoFractional order state equationsfor the control of viscoelastically damped structuresrdquo Journalof Guidance Control and Dynamics vol 14 no 2 pp 304ndash3111991
[2] Z C Li and J S Luo Wavelet Analysis and Its ApplicationElectronic Industrial Publication Beijing China 2005
[3] J H Chen ldquoAnalysis of stability and convergence of numeri-cal approximation for the Riesz fractional reaction-dispersionequationrdquo Journal of Xiamen University (Natural Science) vol46 no 5 pp 616ndash619 2007
[4] D Delbosco and L Rodino ldquoExistence and uniqueness for anonlinear fractional differential equationrdquo Journal of Mathe-matical Analysis and Applications vol 204 no 2 pp 609ndash6251996
[5] Y Ryabov and A Puzenko ldquoA damped oscillations in view ofthe fraction oscillator equationrdquo Physics Review B vol 66 pp184ndash201 2002
[6] A M El-Sayed ldquoNonlinear functional-differential equationsof arbitrary ordersrdquo Nonlinear Analysis Theory Methods ampApplications vol 33 no 2 pp 181ndash186 1998
[7] I L El-Kalla ldquoError estimate of the series solution to a class ofnonlinear fractional differential equationsrdquo Communications inNonlinear Science and Numerical Simulation vol 16 no 3 pp1408ndash1413 2011
[8] Z M Odibat ldquoA study on the convergence of variationaliteration methodrdquo Mathematical and Computer Modelling vol51 no 9-10 pp 1181ndash1192 2010
[9] SMomani ZOdibat andV S Erturk ldquoGeneralized differentialtransform method for solving a space- and time-fractionaldiffusion-wave equationrdquo Physics Letters A vol 370 no 5-6 pp379ndash387 2007
[10] ZOdibat SMomani andV S Erturk ldquoGeneralized differentialtransform method application to differential equations offractional orderrdquo Applied Mathematics and Computation vol197 no 2 pp 467ndash477 2008
[11] Z Odibat and S Momani ldquoA generalized differential transformmethod for linear partial differential equations of fractionalorderrdquo Applied Mathematics Letters vol 21 no 2 pp 194ndash1992008
[12] Y Zhang ldquoA finite difference method for fractional partialdifferential equationrdquo Applied Mathematics and Computationvol 215 no 2 pp 524ndash529 2009
[13] Y X Wang and Q B Fan ldquoThe second kind Chebyshev waveletmethod for solving fractional differential equationsrdquo AppliedMathematics and Computation vol 218 no 17 pp 8592ndash86012012
[14] M X Yi and Y M Chen ldquoHaar wavelet operational matrixmethod for solving fractional partial differential equationsrdquoComputer Modeling in Engineering amp Sciences vol 88 no 3 pp229ndash244 2012
[15] EHDohaAH BhrawyD Baleanu and S S Ezz-Eldien ldquoTheoperational matrix formulation of the Jacobi tau approximationfor space fractional diffusion equationrdquo Advances in DifferenceEquations vol 231 pp 1687ndash1847 2014
[16] M X Yi J Huang and J X Wei ldquoBlock pulse operationalmatrix method for solving fractional partial differential equa-tionrdquo Applied Mathematics and Computation vol 221 pp 121ndash131 2013
Mathematical Problems in Engineering 9
[17] I Podlubny Fractional Differential Equations vol 198 ofMath-ematics in Science and Engineering Academic Press New YorkNY USA 1999
[18] A Saadatmandi and M Dehghan ldquoA new operational matrixfor solving fractional-order differential equationsrdquo ComputersandMathematics with Applications vol 59 no 3 pp 1326ndash13362010
[19] N Liu and E-B Lin ldquoLegendre wavelet method for numericalsolutions of partial differential equationsrdquo Numerical Methodsfor Partial Differential Equations vol 26 no 1 pp 81ndash94 2010
[20] S Nemati and Y Ordokhani ldquoLegendre expansion methodsfor the numerical solution of nonlinear 2D Fredholm integralequations of the second kindrdquo Journal of Applied Mathematicsamp Informatics vol 31 no 5-6 pp 609ndash621 2013
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 Mathematical Problems in Engineering
Table 1 Absolute error for t = 14 s and different values of 119899
119909 n = 2 n = 3 n = 401 03246 32831e minus 015 33582e minus 01602 02351 42734e minus 015 52745e minus 01503 04820 41237e minus 015 56521e minus 01504 03274 57320e minus 015 62742e minus 01605 08231 47381e minus 015 71640e minus 01606 09127 73722e minus 015 32356e minus 01607 12188 63276e minus 015 42745e minus 01508 15181 12374e minus 014 37224e minus 01509 08364 31744e minus 015 88874e minus 015
Table 2 Absolute error for t = 12 s and different values of 119899
119909 n = 2 n = 3 n = 401 02821 42137e minus 015 34325e minus 01602 04375 58711e minus 015 65332e minus 01603 01021 42210e minus 015 52435e minus 01604 02387 31016e minus 016 49722e minus 01605 08277 34762e minus 016 53478e minus 01506 09322 53265e minus 015 13241e minus 01607 13846 47632e minus 015 92371e minus 01608 16654 42346e minus 016 84812e minus 01509 09144 20437e minus 016 63273e minus 016
Table 3 Absolute error for t = 34 s and different values of 119899
119909 n = 2 n = 3 n = 401 03126 50832e minus 016 53281e minus 01602 08978 41845e minus 016 23258e minus 01503 02374 23448e minus 016 40112e minus 01604 02951 32155e minus 016 51223e minus 01605 03327 55518e minus 015 63274e minus 01506 13267 62440e minus 016 57421e minus 01507 08723 35220e minus 016 52871e minus 01608 09229 42301e minus 016 48810e minus 01609 11327 52310e minus 015 61138e minus 016
From Tables 1ndash3 we can see that the absolute error is verysmall when 119899 ge 3 Also when 119899 is fixed the more pointswe take the more accurate numerical solutions we obtainFigures 1ndash3 show the fact that 119899
119909119905is the number of 119909
119894 119905119895
Example 2 Consider the following fractional partial differ-ential equation
119909
12059714
119906 (119909 119905)
12059711990914 +119909
23 12059713
119906 (119909 119905)
12059711990513 = 119891 (119909 119905)
119906 (119909 0) = 119909 (119909 minus 1) (1199092
+ 1) 119906 (0 119905) = 0
(39)
0
2
4
6
002
0406
081
t0
0204
0608
1
x
u(xt)
Figure 1 Numerical solution of 119899119909119905
= 3
0
2
4
6
002
0406
081
t0
0204
0608
1
x
u(xt)
Figure 2 Numerical solution of 119899119909119905
= 6
0
05
1
00204
06081
0
2
4
6
t
x
u(xt)
Figure 3 Exact solution
where
119891 (119909 119905) =
311990523 (55 + 66119905 + 811199053) (minus1 + 119909) 1199092(1 + 119909
2)
110Γ (23)
+
4 (1 + 119905 + 1199052+ 119905
4) 119909
1712[minus385 + 8119909 (55 minus 60119909 + 641199092
)]
1155Γ (34)
(40)
Mathematical Problems in Engineering 7
0020406081 005
1
0
x t
minus02
minus04
minus06
minus08
minus1
minus12
minus14
u(xt)
Figure 4 The numerical solutions for Example 2 when 119899 = 4
00204060810
051
0
xt
minus02
minus04
minus06
minus08
minus1
minus12
minus14
u(xt)
Figure 5 The numerical solutions for Example 2 when 119899 = 5
The exact solution is 119906(119909 119905) = 119909(119909minus 1)(1199092
+1)(1 + 119905 + 1199052
+ 1199054
)The numerical solutions for 119899 = 4 119899 = 5 are displayed inFigures 4 and 5 and the exact solution is shown in Figure 6
Example 3 Consider this equation
11990912
12059713
119906 (119909 119905)
12059711990913
+11990923
12059713
119906 (119909 119905)
12059711990513
= 119891 (119909 119905)
119906 (119909 0) = 1199092
minus 1 119906 (0 119905) = minus1 minus (119905 minus 1)2
minus (119905 minus 1)3
(41)
where119891 (119909 119905)
=
9 (1 + 119905 minus 21199052 + 1199053) 119909
136
5Γ (23)
+
311990523 (minus2 + 3119905) (minus10 + 9119905) (minus1 + 1199092) 119909
23
40Γ (23)
(42)
The exact solution is 119906(119909 119905) = (1199092
minus 1)[1 + (119905 minus 1)2
+ (119905 minus 1)3
]The numerical solutions for 119899 = 3 119899 = 4 are displayed inFigures 7 and 8 The absolute errors are shown in Figures 9and 10 when 119899 = 3 and 119899 = 4
00204060810
05
1
0
xt
minus02
minus04
minus06
minus08
minus1
minus12
minus14
u(xt)
Figure 6 Exact solution for Example 2
0 02 04 06 08 1
0
05
1
0
u(xt)
x
t
minus02
minus04
minus06
minus08
minus1
minus12
minus14
Figure 7 The numerical solutions for Example 3 when 119899 = 3
0 02 04 06 08 1
0
05
1
0
minus02
minus04
minus06
minus08
minus1
minus12
minus14
u(xt)
xt
Figure 8 The numerical solutions for Example 3 when 119899 = 4
From Examples 1ndash3 we can see that the method in thispaper can be effectively used to solve the numerical solutionof fractional partial differential equation with variable coeffi-cients From the above results the absolute errors between
8 Mathematical Problems in Engineering
002
0406
081 0
05
10020406081
12
tx
times10minus14
u(xt)
Figure 9 The absolute errors for Example 3 when 119899 = 3
002
0406
081 0
0204
0608
10
020406081
12
tx
times10minus15
u(xt)
Figure 10 The absolute errors for Example 3 when 119899 = 4
the numerical solutions and the exact solution are rathersmall What is more due to the absolute error in this paperis about 10minus15 the Legendre polynomials method can reachhigher degree of accuracy by comparing the approximationsobtained by block pulse method [16]
7 Conclusion
In this paper we use the Legendre polynomials method tosolve a class of fractional partial differential equations withvariable coefficients The Legendre polynomials operationalmatrix of fractional differentiation is derived from the prop-erty of Legendre polynomials The initial equation is trans-lated into the product of some relevant matrixes which canalso be regarded as the system of linear equations The erroranalysis of Legendre polynomials is also given The numer-ical results show that numerical solutions obtained by ourmethod are in very good agreement with the exact solution
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
References
[1] R L Bagley and R A Calico ldquoFractional order state equationsfor the control of viscoelastically damped structuresrdquo Journalof Guidance Control and Dynamics vol 14 no 2 pp 304ndash3111991
[2] Z C Li and J S Luo Wavelet Analysis and Its ApplicationElectronic Industrial Publication Beijing China 2005
[3] J H Chen ldquoAnalysis of stability and convergence of numeri-cal approximation for the Riesz fractional reaction-dispersionequationrdquo Journal of Xiamen University (Natural Science) vol46 no 5 pp 616ndash619 2007
[4] D Delbosco and L Rodino ldquoExistence and uniqueness for anonlinear fractional differential equationrdquo Journal of Mathe-matical Analysis and Applications vol 204 no 2 pp 609ndash6251996
[5] Y Ryabov and A Puzenko ldquoA damped oscillations in view ofthe fraction oscillator equationrdquo Physics Review B vol 66 pp184ndash201 2002
[6] A M El-Sayed ldquoNonlinear functional-differential equationsof arbitrary ordersrdquo Nonlinear Analysis Theory Methods ampApplications vol 33 no 2 pp 181ndash186 1998
[7] I L El-Kalla ldquoError estimate of the series solution to a class ofnonlinear fractional differential equationsrdquo Communications inNonlinear Science and Numerical Simulation vol 16 no 3 pp1408ndash1413 2011
[8] Z M Odibat ldquoA study on the convergence of variationaliteration methodrdquo Mathematical and Computer Modelling vol51 no 9-10 pp 1181ndash1192 2010
[9] SMomani ZOdibat andV S Erturk ldquoGeneralized differentialtransform method for solving a space- and time-fractionaldiffusion-wave equationrdquo Physics Letters A vol 370 no 5-6 pp379ndash387 2007
[10] ZOdibat SMomani andV S Erturk ldquoGeneralized differentialtransform method application to differential equations offractional orderrdquo Applied Mathematics and Computation vol197 no 2 pp 467ndash477 2008
[11] Z Odibat and S Momani ldquoA generalized differential transformmethod for linear partial differential equations of fractionalorderrdquo Applied Mathematics Letters vol 21 no 2 pp 194ndash1992008
[12] Y Zhang ldquoA finite difference method for fractional partialdifferential equationrdquo Applied Mathematics and Computationvol 215 no 2 pp 524ndash529 2009
[13] Y X Wang and Q B Fan ldquoThe second kind Chebyshev waveletmethod for solving fractional differential equationsrdquo AppliedMathematics and Computation vol 218 no 17 pp 8592ndash86012012
[14] M X Yi and Y M Chen ldquoHaar wavelet operational matrixmethod for solving fractional partial differential equationsrdquoComputer Modeling in Engineering amp Sciences vol 88 no 3 pp229ndash244 2012
[15] EHDohaAH BhrawyD Baleanu and S S Ezz-Eldien ldquoTheoperational matrix formulation of the Jacobi tau approximationfor space fractional diffusion equationrdquo Advances in DifferenceEquations vol 231 pp 1687ndash1847 2014
[16] M X Yi J Huang and J X Wei ldquoBlock pulse operationalmatrix method for solving fractional partial differential equa-tionrdquo Applied Mathematics and Computation vol 221 pp 121ndash131 2013
Mathematical Problems in Engineering 9
[17] I Podlubny Fractional Differential Equations vol 198 ofMath-ematics in Science and Engineering Academic Press New YorkNY USA 1999
[18] A Saadatmandi and M Dehghan ldquoA new operational matrixfor solving fractional-order differential equationsrdquo ComputersandMathematics with Applications vol 59 no 3 pp 1326ndash13362010
[19] N Liu and E-B Lin ldquoLegendre wavelet method for numericalsolutions of partial differential equationsrdquo Numerical Methodsfor Partial Differential Equations vol 26 no 1 pp 81ndash94 2010
[20] S Nemati and Y Ordokhani ldquoLegendre expansion methodsfor the numerical solution of nonlinear 2D Fredholm integralequations of the second kindrdquo Journal of Applied Mathematicsamp Informatics vol 31 no 5-6 pp 609ndash621 2013
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
Mathematical Problems in Engineering 7
0020406081 005
1
0
x t
minus02
minus04
minus06
minus08
minus1
minus12
minus14
u(xt)
Figure 4 The numerical solutions for Example 2 when 119899 = 4
00204060810
051
0
xt
minus02
minus04
minus06
minus08
minus1
minus12
minus14
u(xt)
Figure 5 The numerical solutions for Example 2 when 119899 = 5
The exact solution is 119906(119909 119905) = 119909(119909minus 1)(1199092
+1)(1 + 119905 + 1199052
+ 1199054
)The numerical solutions for 119899 = 4 119899 = 5 are displayed inFigures 4 and 5 and the exact solution is shown in Figure 6
Example 3 Consider this equation
11990912
12059713
119906 (119909 119905)
12059711990913
+11990923
12059713
119906 (119909 119905)
12059711990513
= 119891 (119909 119905)
119906 (119909 0) = 1199092
minus 1 119906 (0 119905) = minus1 minus (119905 minus 1)2
minus (119905 minus 1)3
(41)
where119891 (119909 119905)
=
9 (1 + 119905 minus 21199052 + 1199053) 119909
136
5Γ (23)
+
311990523 (minus2 + 3119905) (minus10 + 9119905) (minus1 + 1199092) 119909
23
40Γ (23)
(42)
The exact solution is 119906(119909 119905) = (1199092
minus 1)[1 + (119905 minus 1)2
+ (119905 minus 1)3
]The numerical solutions for 119899 = 3 119899 = 4 are displayed inFigures 7 and 8 The absolute errors are shown in Figures 9and 10 when 119899 = 3 and 119899 = 4
00204060810
05
1
0
xt
minus02
minus04
minus06
minus08
minus1
minus12
minus14
u(xt)
Figure 6 Exact solution for Example 2
0 02 04 06 08 1
0
05
1
0
u(xt)
x
t
minus02
minus04
minus06
minus08
minus1
minus12
minus14
Figure 7 The numerical solutions for Example 3 when 119899 = 3
0 02 04 06 08 1
0
05
1
0
minus02
minus04
minus06
minus08
minus1
minus12
minus14
u(xt)
xt
Figure 8 The numerical solutions for Example 3 when 119899 = 4
From Examples 1ndash3 we can see that the method in thispaper can be effectively used to solve the numerical solutionof fractional partial differential equation with variable coeffi-cients From the above results the absolute errors between
8 Mathematical Problems in Engineering
002
0406
081 0
05
10020406081
12
tx
times10minus14
u(xt)
Figure 9 The absolute errors for Example 3 when 119899 = 3
002
0406
081 0
0204
0608
10
020406081
12
tx
times10minus15
u(xt)
Figure 10 The absolute errors for Example 3 when 119899 = 4
the numerical solutions and the exact solution are rathersmall What is more due to the absolute error in this paperis about 10minus15 the Legendre polynomials method can reachhigher degree of accuracy by comparing the approximationsobtained by block pulse method [16]
7 Conclusion
In this paper we use the Legendre polynomials method tosolve a class of fractional partial differential equations withvariable coefficients The Legendre polynomials operationalmatrix of fractional differentiation is derived from the prop-erty of Legendre polynomials The initial equation is trans-lated into the product of some relevant matrixes which canalso be regarded as the system of linear equations The erroranalysis of Legendre polynomials is also given The numer-ical results show that numerical solutions obtained by ourmethod are in very good agreement with the exact solution
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
References
[1] R L Bagley and R A Calico ldquoFractional order state equationsfor the control of viscoelastically damped structuresrdquo Journalof Guidance Control and Dynamics vol 14 no 2 pp 304ndash3111991
[2] Z C Li and J S Luo Wavelet Analysis and Its ApplicationElectronic Industrial Publication Beijing China 2005
[3] J H Chen ldquoAnalysis of stability and convergence of numeri-cal approximation for the Riesz fractional reaction-dispersionequationrdquo Journal of Xiamen University (Natural Science) vol46 no 5 pp 616ndash619 2007
[4] D Delbosco and L Rodino ldquoExistence and uniqueness for anonlinear fractional differential equationrdquo Journal of Mathe-matical Analysis and Applications vol 204 no 2 pp 609ndash6251996
[5] Y Ryabov and A Puzenko ldquoA damped oscillations in view ofthe fraction oscillator equationrdquo Physics Review B vol 66 pp184ndash201 2002
[6] A M El-Sayed ldquoNonlinear functional-differential equationsof arbitrary ordersrdquo Nonlinear Analysis Theory Methods ampApplications vol 33 no 2 pp 181ndash186 1998
[7] I L El-Kalla ldquoError estimate of the series solution to a class ofnonlinear fractional differential equationsrdquo Communications inNonlinear Science and Numerical Simulation vol 16 no 3 pp1408ndash1413 2011
[8] Z M Odibat ldquoA study on the convergence of variationaliteration methodrdquo Mathematical and Computer Modelling vol51 no 9-10 pp 1181ndash1192 2010
[9] SMomani ZOdibat andV S Erturk ldquoGeneralized differentialtransform method for solving a space- and time-fractionaldiffusion-wave equationrdquo Physics Letters A vol 370 no 5-6 pp379ndash387 2007
[10] ZOdibat SMomani andV S Erturk ldquoGeneralized differentialtransform method application to differential equations offractional orderrdquo Applied Mathematics and Computation vol197 no 2 pp 467ndash477 2008
[11] Z Odibat and S Momani ldquoA generalized differential transformmethod for linear partial differential equations of fractionalorderrdquo Applied Mathematics Letters vol 21 no 2 pp 194ndash1992008
[12] Y Zhang ldquoA finite difference method for fractional partialdifferential equationrdquo Applied Mathematics and Computationvol 215 no 2 pp 524ndash529 2009
[13] Y X Wang and Q B Fan ldquoThe second kind Chebyshev waveletmethod for solving fractional differential equationsrdquo AppliedMathematics and Computation vol 218 no 17 pp 8592ndash86012012
[14] M X Yi and Y M Chen ldquoHaar wavelet operational matrixmethod for solving fractional partial differential equationsrdquoComputer Modeling in Engineering amp Sciences vol 88 no 3 pp229ndash244 2012
[15] EHDohaAH BhrawyD Baleanu and S S Ezz-Eldien ldquoTheoperational matrix formulation of the Jacobi tau approximationfor space fractional diffusion equationrdquo Advances in DifferenceEquations vol 231 pp 1687ndash1847 2014
[16] M X Yi J Huang and J X Wei ldquoBlock pulse operationalmatrix method for solving fractional partial differential equa-tionrdquo Applied Mathematics and Computation vol 221 pp 121ndash131 2013
Mathematical Problems in Engineering 9
[17] I Podlubny Fractional Differential Equations vol 198 ofMath-ematics in Science and Engineering Academic Press New YorkNY USA 1999
[18] A Saadatmandi and M Dehghan ldquoA new operational matrixfor solving fractional-order differential equationsrdquo ComputersandMathematics with Applications vol 59 no 3 pp 1326ndash13362010
[19] N Liu and E-B Lin ldquoLegendre wavelet method for numericalsolutions of partial differential equationsrdquo Numerical Methodsfor Partial Differential Equations vol 26 no 1 pp 81ndash94 2010
[20] S Nemati and Y Ordokhani ldquoLegendre expansion methodsfor the numerical solution of nonlinear 2D Fredholm integralequations of the second kindrdquo Journal of Applied Mathematicsamp Informatics vol 31 no 5-6 pp 609ndash621 2013
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
8 Mathematical Problems in Engineering
002
0406
081 0
05
10020406081
12
tx
times10minus14
u(xt)
Figure 9 The absolute errors for Example 3 when 119899 = 3
002
0406
081 0
0204
0608
10
020406081
12
tx
times10minus15
u(xt)
Figure 10 The absolute errors for Example 3 when 119899 = 4
the numerical solutions and the exact solution are rathersmall What is more due to the absolute error in this paperis about 10minus15 the Legendre polynomials method can reachhigher degree of accuracy by comparing the approximationsobtained by block pulse method [16]
7 Conclusion
In this paper we use the Legendre polynomials method tosolve a class of fractional partial differential equations withvariable coefficients The Legendre polynomials operationalmatrix of fractional differentiation is derived from the prop-erty of Legendre polynomials The initial equation is trans-lated into the product of some relevant matrixes which canalso be regarded as the system of linear equations The erroranalysis of Legendre polynomials is also given The numer-ical results show that numerical solutions obtained by ourmethod are in very good agreement with the exact solution
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
References
[1] R L Bagley and R A Calico ldquoFractional order state equationsfor the control of viscoelastically damped structuresrdquo Journalof Guidance Control and Dynamics vol 14 no 2 pp 304ndash3111991
[2] Z C Li and J S Luo Wavelet Analysis and Its ApplicationElectronic Industrial Publication Beijing China 2005
[3] J H Chen ldquoAnalysis of stability and convergence of numeri-cal approximation for the Riesz fractional reaction-dispersionequationrdquo Journal of Xiamen University (Natural Science) vol46 no 5 pp 616ndash619 2007
[4] D Delbosco and L Rodino ldquoExistence and uniqueness for anonlinear fractional differential equationrdquo Journal of Mathe-matical Analysis and Applications vol 204 no 2 pp 609ndash6251996
[5] Y Ryabov and A Puzenko ldquoA damped oscillations in view ofthe fraction oscillator equationrdquo Physics Review B vol 66 pp184ndash201 2002
[6] A M El-Sayed ldquoNonlinear functional-differential equationsof arbitrary ordersrdquo Nonlinear Analysis Theory Methods ampApplications vol 33 no 2 pp 181ndash186 1998
[7] I L El-Kalla ldquoError estimate of the series solution to a class ofnonlinear fractional differential equationsrdquo Communications inNonlinear Science and Numerical Simulation vol 16 no 3 pp1408ndash1413 2011
[8] Z M Odibat ldquoA study on the convergence of variationaliteration methodrdquo Mathematical and Computer Modelling vol51 no 9-10 pp 1181ndash1192 2010
[9] SMomani ZOdibat andV S Erturk ldquoGeneralized differentialtransform method for solving a space- and time-fractionaldiffusion-wave equationrdquo Physics Letters A vol 370 no 5-6 pp379ndash387 2007
[10] ZOdibat SMomani andV S Erturk ldquoGeneralized differentialtransform method application to differential equations offractional orderrdquo Applied Mathematics and Computation vol197 no 2 pp 467ndash477 2008
[11] Z Odibat and S Momani ldquoA generalized differential transformmethod for linear partial differential equations of fractionalorderrdquo Applied Mathematics Letters vol 21 no 2 pp 194ndash1992008
[12] Y Zhang ldquoA finite difference method for fractional partialdifferential equationrdquo Applied Mathematics and Computationvol 215 no 2 pp 524ndash529 2009
[13] Y X Wang and Q B Fan ldquoThe second kind Chebyshev waveletmethod for solving fractional differential equationsrdquo AppliedMathematics and Computation vol 218 no 17 pp 8592ndash86012012
[14] M X Yi and Y M Chen ldquoHaar wavelet operational matrixmethod for solving fractional partial differential equationsrdquoComputer Modeling in Engineering amp Sciences vol 88 no 3 pp229ndash244 2012
[15] EHDohaAH BhrawyD Baleanu and S S Ezz-Eldien ldquoTheoperational matrix formulation of the Jacobi tau approximationfor space fractional diffusion equationrdquo Advances in DifferenceEquations vol 231 pp 1687ndash1847 2014
[16] M X Yi J Huang and J X Wei ldquoBlock pulse operationalmatrix method for solving fractional partial differential equa-tionrdquo Applied Mathematics and Computation vol 221 pp 121ndash131 2013
Mathematical Problems in Engineering 9
[17] I Podlubny Fractional Differential Equations vol 198 ofMath-ematics in Science and Engineering Academic Press New YorkNY USA 1999
[18] A Saadatmandi and M Dehghan ldquoA new operational matrixfor solving fractional-order differential equationsrdquo ComputersandMathematics with Applications vol 59 no 3 pp 1326ndash13362010
[19] N Liu and E-B Lin ldquoLegendre wavelet method for numericalsolutions of partial differential equationsrdquo Numerical Methodsfor Partial Differential Equations vol 26 no 1 pp 81ndash94 2010
[20] S Nemati and Y Ordokhani ldquoLegendre expansion methodsfor the numerical solution of nonlinear 2D Fredholm integralequations of the second kindrdquo Journal of Applied Mathematicsamp Informatics vol 31 no 5-6 pp 609ndash621 2013
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
Mathematical Problems in Engineering 9
[17] I Podlubny Fractional Differential Equations vol 198 ofMath-ematics in Science and Engineering Academic Press New YorkNY USA 1999
[18] A Saadatmandi and M Dehghan ldquoA new operational matrixfor solving fractional-order differential equationsrdquo ComputersandMathematics with Applications vol 59 no 3 pp 1326ndash13362010
[19] N Liu and E-B Lin ldquoLegendre wavelet method for numericalsolutions of partial differential equationsrdquo Numerical Methodsfor Partial Differential Equations vol 26 no 1 pp 81ndash94 2010
[20] S Nemati and Y Ordokhani ldquoLegendre expansion methodsfor the numerical solution of nonlinear 2D Fredholm integralequations of the second kindrdquo Journal of Applied Mathematicsamp Informatics vol 31 no 5-6 pp 609ndash621 2013
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