Fractional One-Dimensional Transport Equation ...

Proceedings of the ASME 2009 International Design Engineering Technical Conferences & Computers and Information in Engineering Conference Proceedings of the ASME 2009 International Design Engineering Technical Conferences IDETC/CIE 2009 & Computers and Information in Engineering Conference August 30 - September 2, 2009, San Diego, California, USA IDETC/CIE 2009 August 30-September 2, 2009, San Diego, USA

DETC2009-86878 DETC2009-86878

FRACTIONAL ONE-DIMENSIONAL TRANSPORT EQUATION WITHIN SPECTRAL METHOD COMBINED WITH MODIFIED ADOMIAN DECOMPOSITION METHOD

Dumitru Baleanu Department of Mathematics and Computer Sciences Faculty of Art and Sciences Cankaya University 06530 Balgat, Ankara, Turkey, Email: [email protected] and Institute of Space Sciences Magurele-Bucharest,P.O.BOX, MG-23, R 76900, Romania Email: [email protected] Abdelouahab Kadem L.M.F.N., Mathematics Department, University of Setif, Algeria Email:[email protected]

ABSTRACT In this paper the Chebyshev polynomials technique combined with the modified Adomian decomposition method were applied to solve analytically the fractional transport equation in one-dimensional plane geometry.

initiated by Riewe [15] and continued by several authors (see the following references [16–20] as well as the references therein ) lead us to some new important results in the theory of control and physics. Recently, a new approach of solving the fractional transport equation was proposed [21]. A key role was playing by the Chebyshev polynomials [21–23]. The approach is based on expansion of the angular flux in a truncated series of Chebyshev polynomials in the angular variable [24]. By replacing this development to the transport equation we will get a first-order linear differential system which can be solved for the spatial function coefficients by using the Sumudu transform [25]. The inversion of the transformed coefficients is obtained by using the Trzaska’s method [26] and the Heaviside expansion technique. The main aim of this work is to solve the one-dimensional fractional transport equation in a finite domain by using the combined use of the spectral method and modified Adomian decom-

1

INTRODUCTION Fractional calculus is deeply related to the problem of the extension of meaning. This type of calculus deals with the fractional derivatives and integral of any order [1–4]. The applications of fractional calculus are very diverse and during the last decades important results were reported in several branches of science and engineering (see for example Refs. [5– 13] and the references therein). Although the physical meaning of the fractional derivatives is not clearly understood [14] the fractional variational principles 1

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position method. The paper has been organized as follows: Section 2 contains some preliminaries on fractional calculus. Section 3 presents briefly the review of the standard modified Adomian decomposition. Section 4 deals with the fractional transport equation in one dimensional case. We applied the modified Adomian decomposition to solve this equation. Finally, Section 5 is dedicated to our conclusions.

µ

I µC D0+ f (t) = f (t)−

m−1

∑

f (k) (0+ )

k=0

and C µ D0+

f (t) = Dµ

m−1

f (t) −

∑

k=0

BASIC TOOLS In the following we review briefly some basic definitions and basic results within the fractional calculus are [1–4]. A real function f (x), x > 0 is said to be in the space Cα , α ∈ R if there exists a real number p (p > α), such that f (x) = x p f1 (x) where f1 (x) = C [0, ∞) . Thus, Cα ⊂ Cβ if β ≤ α. A function f (x), x > 0 is said to be in space Cαm , m ∈ N ∪{0}, (m) if f ∈ Cα . The (left-hand sided) Riemann-Liouville fractional integral of the order µ ≥ 0 had the following form 1 I f (t) = Γ(µ)

0

(t − τ)

µ−1

f (τ)dτ, µ > 0,t > 0,

I 0 f (t) = f (t).

d m m−µ I f (t), m − 1 < µ ≤ m, m ∈ N. dt m

(1)

Dµc f (t) =

dm dt m

α > β, α = β, α < β,

(8)

β

D0+ Dm f (t) = Dβ+m f (t),

m = 0, 1, 2, ..., n − 1 < β < n. (9)

3 THE METHOD In the following a brief review of the standard Adomian decomposition method is presented (see for more details Refs. [27, 28]). The starting point is the following differential equation,

(2)

Lu + Ru + Nu = g(x),

(10)

where N denotes a non-linear operator, L is the highest-order derivative which is assumed in this paper to be invertible, R denotes a linear differential operator of less order than L and g is the source term. By applying the operator L−1 formally to the following expression

(3)

In addition of that, the (left sided) Caputo fractional derivam , m ∈ N ∪ {0} of order α > 0, is defined as tive of f , f ∈ C−1 i (h I m−µ f (m) (t) , m − 1 < µ ≤ m, m ∈ N,

, m−1 < µ ≤ m, m ∈ N.

and

As a result, the operator I µ , µ > 0 is well defined on the functional space Cα , α ≥ −1. The (left-hand sided) Riemann-Liouville fractional derivative of the order µ > 0 is defined by Dµ f (t) =

!

 α−β f (t) if I f (t) if Dβ I α f (t) =  β−α D f (t) if

C µ

tk f (k) (0+ ) k!

(7) Finally, we enlist two more basic properties of the fractional derivatives and integrals, namely

2

Z t

tk , m−1 < µ ≤ m, m ∈ N, (6) k!

Lu = g − Ru − Nu,

(4)

(11)

f (t)µ = m. and by using the given conditions we get the following

Some of the basic properties of the defined fractional derivatives and integrals are given below, namely

I µt γ =

Γ(γ + 1) γ+µ t , µ > 0, γ > −1,t > 0, Γ(γ + µ + 1)

u = f − L−1 (Ru) − L−1 (Nu),

(12)

where the function f denotes the terms arising from integrating the source term g, and from the use of the given conditions.

(5) 2


+∞ The next step is to write Nu = ∑+∞ n=0 An and u = ∑n=0 un , where the components of An are the so-called Adomian polynomials. We observe that for each i,Ai depends on u0, u1,..., ui only. By considering (12), we define u0 , u1 , ·, un as given below

u0 = L−1 (g), u1 = −L−1 (Ru0 ) − L−1 (Nu0 ),  u2 = −L−1 (Ru1 ) − L−1 (Nu1 ),  

By making use of (12,13) and (21), we obtain the following  u0 = L−1 (a0 T0 (x) + a1 T1 (x) + a2 T2 (x) + ... + aλ Tλ (x)),    u1 = −L−1 (Ru0 ) − L−1 (Nu0 ), u2 = −L−1 (Ru1 ) − L−1 (Nu1 ),    ··· (22)

(13) 4

ONE-DIMENSIONAL FRACTIONAL TRANSPORT EQUATION In the following we consider a one-dimensional fractional transport equation and we will solve it by using the modified Adomian method decomposition. This equation is obtained by the classical one by replacing the classical derivative with Caputo derivative. As a result, the corresponding fractional equation is given by

or in the general case we obtain uk+1 = −L−1 (Ruk ) − L−1 (Nuk ), k ≥ 0.

(14)

If the series converges, then we observe that u = lim ΨM (x),

β

(15)

M→∞

µC D0+ Ψ(x, y, µ, φ) + σt Ψ(x, µ) = +

where ΨM (x) = Therefore, an expression for Ai is required. Specific algorithms were developed in [29–31] to formulate the Adomian polynomials. The algorithm given below,namely ∑M i=0 ui .

A0 = F(u0 ), A1 = u1 F 0 (u0 ),

(16) (17)

1 A2 = u2 F 0 (u0 ) + u21 F 00 (u0 ), 2

(18)

A3 = u3 F 0 (u0 ) + u1 u2 F 00 (u0 ) + ···

1 3 000 u F (u0 ), 3! 1

−1

σs (µ, µ0 )Ψ(x, µ0 )dµ0

q(x) , 2

(23)

such that σs (µ, µ0 ) =

∞

2k + 1 σsk Pk (µ)Pk (µ0 ). 2 k=0

∑

(24)

We consider the case 0 ≤ x ≤ a , µ ∈ [−1, 1] and 0 < β ≤ 1. The boundary conditions are given below as below Ψ(x = 0, −µ) = 0,

(25)

Ψ(x = a, µ) = f (µ),

(26)

(19) (20)

can be used to construct the Adomian polynomials in the case when F(u) = N(u). In order to perform the Adomian decomposition method for an arbitrary given natural number λ, g(x) is expressed in Chebyshev series as given below

where f (µ) is the prescribed incident flux at x = a; Ψ(x, µ) is the angular flux in the µ direction σt is the total cross section; σsl , with l = 0, ..., L are the components of the differential scattering cross section and Pk (µ) are the Legendre polynomials of degree k. Theorem. Consider the integro-differential equation subject to boundary conditions and then the function Ψ(x, µ) satisfy the follow first-order differential equation system for the components Yn (x) [21]

λ

g(x) = ∑ ai Ti (x).

Z 1

(21)

i=0

β

Here Ti (x) denotes the first kind of the orthogonal Chebyshev polynomials.

∑Nn=0 αC D0+ Yn + 3

L 2l + 1 2 N 3 σt π Yn (x) = ∑ αs,l ∑ αn,l Yn (x) 2 − δm,0 2 n=0 l=0


+

q(x) , 2

Proof We can easily observe that we have the following relation

(27)

where α1n,m = α1n,m α2n,l

Z 1

Tm (µ) := µTn (µ) p dµ, −1 1 − µ2 −1

α3n,l : =

Tn (µ)Pl (µ)dµ,

Z 1 Tm (µ)Pl (µ) −1

p dµ. 1 − µ2

(36)

(28) For k = 2 and by using the multiplication of the Chebyshev and the Legendre recurrence formulas we obtain

Z 1

:=

πδ|n−m| . 2(2 − δn+m,1 )

(29)

2l+1 2l+2

(30)

[Pl (µ)Tn+1 (µ) + Pl (µ)Tn−1 (µ)] l − Pl−1 (µ) [Tn+1 (µ) + Tn−1 (µ)] . 2µ(l + 1)

and Yn (x) are the coefficients of the expansions of the Ψ(x, µ). To be prepared for the proof we need the following result.

(37)

By using the well known relation

Let Tn+1 (µ) + Tn−1 (µ) = 2µTn (µ) Tn+1 (x) − 2xTn (x) + Tn−1 (x) = 0

(31) and after doing some algebraic manipulations and by using the integration over µ ∈ [−1, 1] on the resulting equation we obtain the following

and Pl+1 (x) = 2xPl (x) − Pl−1 (x) − [xPl (x) − Pl−1 (x)] /(l + 1) (32)

α2n,l+1 :=

be the recurrence relations for the Chebyshev and the Legendre polynomials respectively. For l > 2 and k = 2, 3 we obtain αkn,l+1 :=

α3n,l

(38)

i l 2l + 1 h k αn+1,l + αkn−1,l − αk . 2l + 2 l + 1 n, j−1

α2n,l =

0

if

2 (1+l 2 )−n2

n+l odd if n+l even

1−µ

µ ∈ [−1, 1] before getting the desired result. In the following we give the proof of the above mentioned Theorem [21].

(33)

Proof. We expand the angular flux in the variable µ in term of (x)Tn (µ) the Chebyshev polynomials we obtain Ψ(x, µ) = ∑Nn=0 Yn√ , 2 1−µ

such that N = 0, 2, 4, ..., where the expansions coefficients should be determined. Tn (µ) denote the Chebyshev polynomials of order n which are orthogonal √ in the interval [−1, 1] with respect to the weight ω(t) = 1/ 1 − t 2 . After replacing this expression into (23) we obtain

(34)

and α3n,l =

πδn,l . 2 − δl,0

(39)

The case k = 3 can be treated similarly but in this case we multiply the resulting expression by √ 1 2 and integrate over

Hence, in particular for l = 0 and 1 the coefficients α2n,l and assume the values (

2l + 1 2 l αn+1,l + α2n−1,l − α2 . 2l + 2 l + 1 n, j−1

(35) n o T (µ) n β µC D0+ Yn (x) + σt Yn (x) p = 1 − µ2 Z 1 N Tm (µ0 ) 2l + 1 αs,l Pl (µ) ∑ Yn (x) Pl (µ0 ) p dµ0 ∑Ll=0 2 −1 1 − µ02 n=0 q(x) , (40) + 2

∑Nn=0 We noticed that the Chebyshev polynomials are important in approximation theory because the roots of the Chebyshev polynomials of the first kind are utilized as nodes in polynomial interpolation.

4


Here, the coefficients have the following expressions

using the orthogonality of the Chebyshev polynomials, multiply the by Tm (µ), considering m = 0, ..., N, and integrated in the µ variable in the interval [−1, 1]. Thus, we get the following firstorder linear differential equation system for the spatial component Yn (x) σt π Yn (x) 2 − δm,0 L 2l + 1 2 N 3 q(x) = ∑ αs,l ∑ αn,l Yn (x) + , 2 2 n=0 l=0

1 a0 = 2π

Z 1

q(x) √ dx −1 1 − x2

(49)

and

β

∑Nn=0 α1C n,m D0+ Yn (x) +

ai = (41)

1 π

Z 1

q(x) √ dx. −1 1 − x2

(50)

In our case N = 0, therefore the Adomian polynomials Ai = 0. As β a result we have L−1 = I0+ . Finally, we obtain the expressions for Yi , i = 1, · · · N as given below

where α1n,m :=

α2n,l

Z 1 −1

Tm (µ) dµ µTn (µ) p 1 − µ2

(42)

1 β I Y0 (x) = I0+ ( ∑ ai Ti (x)), 2 i=0

(51)

Z 1

:= −1

α3n,l :=

Tn (µ)Pl (µ)dµ

Z 1 Tm (µ)Pl (µ) −1

p dµ. 1 − µ2

(43)

β

Y1 = −I0+ (RY0 (x)),

β

(44)

Y2 = −I0+ (RY1 (x)),

with δn,m denoting the delta of Kronecker. Here the coefficients α2n,l and α3n,l are evaluated by the multiplication of the Chebyshev and the Legendre recurrence formulas and integration of the resulting equation. At this stage, our aim is to solve the above mentioned fractional differential equation by using the modified Adomian decomposition method. We obtain that Lu + Ru + Nu = g(x),

β

Y3 = −I0+ (RY2 (x)).

(52)

(53)

(54)

In the general case we obtain the following expression for Yn , namely β

Yn = −I0+ (RYn−1 (x)).

(55)

(45) 5

CONCLUSIONS The transport equation in one or higher dimensions have various applications in several fields. The fractional generalization of this type of equation presents its own interest mainly because of the non-locality appearing within the fractional derivative. In this paper we have analyzed the fractional onedimensional transport equation within Caputo derivative and we solve it analytically by using the modified Adomian decomposition method. This method is appropriate to solve the investigated problem because the term g(x) is considered as a linear combination of the Chebyshev polynomials. As a results the combined technique is easy to apply and it leads to the final results faster than some other standard methods.

where β

L =C D0+ ,

R=

L σt π 2l + 1 2 N 3 −∑ αs,l ∑ αn,l 2 − δm,0 l=0 2 n=0

(46)

(47)

and g(x) =

q(x) 1 I ' ∑ ai Ti (x). 2 2 i=0

(48) 5


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