AAAAbstract and Applied Analysis1687-04091085-3375Hindawi Publishing Corporation70416810.1155/2010/704168704168Research ArticleNote on the Solution of Transport Equation by Tau Method and Walsh FunctionsKademAbdelouahab1KilicmanAdem2GigaYoshikazu1L.M.F.N Mathematics DepartmentUniversity of Setif, Setif 19000Algeriauniv-setif.dz2Department of Mathematics and Institute for Mathematical ResearchUniversity Putra Malaysia (UPM)Serdang, Selangor 43400Malaysiaupm.edu.my201030122010201009112010271220102010Copyright © 2010 Abdelouahab Kadem and Adem Kilicman.This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

We consider the combined Walsh function for the three-dimensional case. A method for the solution of the neutron transport equation in three-dimensional case by using the Walsh function, Chebyshev polynomials, and the Legendre polynomials are considered. We also present Tau method, and it was proved that it is a good approximate to exact solutions. This method is based on expansion of the angular flux in a truncated series of Walsh function in the angular variable. The main characteristic of this technique is that it reduces the problems to those of solving a system of algebraic equations; thus, it is greatly simplifying the problem.

1. Introduction

The Walsh functions have many properties similar to those of the trigonometric functions. For example, they form a complete, total collection of functions with respect to the space of square Lebesgue integrable functions. However, they are simpler in structure to the trigonometric functions because they take only the values 1 and −1. They may be expressed as linear combinations of the Haar functions , so many proofs about the Haar functions carry over to the Walsh system easily. Moreover, the Walsh functions are Haar wavelet packets. For a good account of the properties of the Haar wavelets and other wavelets, see . We use the ordering of the Walsh functions due to Paley [3, 4]. Any function fL2[0,1] can be expanded as a series of Walsh functionsf(x)=i=0ciWi(x),whereci=01f(x)Wi(x). In , Fine discovered an important property of the Walsh Fourier series: the m=2nth partial sum of the Walsh series of a function f is piecewise constant, equal to the L1 mean of f, on each subinterval [(i-1)/m,i/m]. For this reason, Walsh series in applications are always truncated to m=2n terms. In this case, the coefficients ci of the Walsh (-Fourier) series are given by ci=j=0m-11mWijfj, where fj is the average value of the function f(x) in the jth interval of width 1/m in the interval (0,1), and Wij is the value of the ith Walsh function in the jth subinterval. The order m Walsh matrix, 𝒲m, has elements Wij.

Let f(x) have a Walsh series with coefficients ci and its integral from 0 to x have a Walsh series with coefficients bi: 0xf(t)dt=i=0biWi(x). If we truncate to m=2n terms and use the obvious vector notation, then integration is performed by matrix multiplication b=PmTc, wherePmT=[Pm/212mIm/2-12mIm/2Om/2],P2T=[1214-140], and Im is the unit matrix, Om is the zero matrix (of order m), see .

2. The Three-Dimensional Spectral Solution

In the literature there several works on driving a suitable model for the transport equation in 2 and 3-dimensional case as well as in cylindrical domain, for example, see , and by using the eigenvalue error estimates for two-dimensional neutron transport, see , by applying the finite element method in an infinite cylindrical domain, see , similarly by using Chebyshev spectral-SN method, see , and the discrete ordinates in the infinite cylindrical domain, see .

In this paper, we consider combined Walsh function with the Sumudu transform in order to extend the transport problem for the three-dimensional case by following the similar method that was proposed in . This method is based on expansion of the angular flux in a truncated series of Walsh function in the angular variable. By replacing this development in the transport equation, this will result a first-order linear differential system. First of all we consider the three-dimensional linear, steady state, transport equation which is given by μxΨ(x,μ,θ)+1-μ2(cosθyΨ(x,μ,θ)+sinθzΨ(x,μ,θ))+σtΨ(x,μ,θ)=-1102πσs(μ,θμ,θ)Ψ(x,μ,θ)dθdμ+S(x,μ,θ), where we assume that the spatial variable x:=(x,y,z) varies in the cubic domain Ω:={(x,y,z):-1x,y,z1}, and Ψ(x,μ,θ):=Ψ(x,y,z,μ,θ) is the angular flux in the direction defined by μ[-1,1] and θ[0,2π]. σt and σs denote the total and the differential cross section, respectively, σs(μ,ϕμ,ϕ) describes the scattering from an assumed pre-collision angular coordinates (μ,θ) to a postcollision coordinates (μ,θ) and S is the source term. See  for further details.

Note that, in the case of one-speed neutron transport equation; taking the angular variable in a disc, this problem will corresponds to a three dimensional case with all functions being constant in the azimuthal direction of the z variable. In this way the actual spatial domain may be assumed to be a cylinder with the cross-section Ω and the axial symmetry in z. Then D will correspond to the projection of the points on the unit sphere (the “speed”) onto the unit disc (which coincides with D). See  for the details.

Given the functions f1(y,z,μ,ϕ), f2(x,z,μ,ϕ), and f3(x,y,μ,ϕ) describing the incident flux, we seek for a solution of (2.1) subject to the following boundary conditions.

For the boundary terms in x, for 0θ2π, letΨ(x=±1,y,z,μ,θ)={f1(y,z,μ,θ),x=-1,0<μ1,0,x=1,-1μ<0. For the boundary terms in y and for -1μ<1,Ψ(x,y=±1,z,μ,θ)={f2(x,z,μ,θ),y=-1,0<cosθ1,0,y=1,-1cosθ<0. Finally, for the boundary terms in z, for -1μ<1,Ψ(x,y,z=±1,μ,θ)={f3(x,y,μ,θ),z=-1,0θ<π,0,z=1,π<θ2π.

Theorem 2.1.

Consider the integrodifferential equation (2.1) under the boundary conditions (2.2), (2.3) and (2.4), then the function Ψ(x,y,z,μ,θ) satisfy the following first-order linear differential equation system for the spatial component Ψi,j(x,μ,θ)μΨi,jx(x,μ,θ)+σtΨi,j(x,μ,θ)  =G  i,j(x;μ,θ)-11σs(μ,θμ,θ)Ψi,j(x,μ,θ)dθdμ, with the boundary conditions Ψi,j(-1,μ,η)=f1i,j(μ,θ), where f1i,j(μ,θ)=4π2-11Ti(y)Rj(z)(1-y2)(1-z2)f1(y,z,μ,θ)dzdy,Ψi,j(1,-μ,θ)=0,Gi,j(x;μ,θ)=Si,j(x,μ,θ)-1-μ2[cosθk=i+1IAikΨk,j(x,μ,θ)+sinθl=j+1JBjlΨi,l(x,μ,θ)], with Si,j(x,μ,θ)=4π2-11Ti(y)Rj(z)(1-y2)(1-z2)S(x,μ,θ)dzdy,Aik=2π-11ddy(Tk(y))Ti(y)1-y2dy,  Bjl=2π-11ddy(Rl(y))Rj(z)1-z2dz.

Proof.

Expanding the angular flux Ψ(x,y,z,μ,ϕ) in a truncated series of Chebyshev polynomials Ti(y) and Rj(z) leads to Ψ(x,y,z,μ,θ)=i=0Ij=0JΨi,j(x,μ,θ)Ti(y)Rj(z). We insert Ψ(x,y,z,μ,θ) given by (2.9) into the boundary condition in (2.3), for y=±1. Multiplying the resulting expressions by Rj(z)/1-z2 and integrating over z, we get the components Ψ0,j(x,μ,θ) for j=0,,J, Ψ0,j(x,μ,θ)=f2j(x,μ,θ)-i=1I(-1)jΨi,j(x,μ,θ),0<cosθ1,Ψ0,j(x,μ,θ)=-i=1IΨi,j(x,μ,θ),-1cosθ<0. Similarly, we substitute Ψ(x,y,z,μ,θ) from (2.9) into the boundary conditions for z=±1, multiply the resulting expression by Ti(y)/1-y2, i=0,,I and integrating over y, to define the components Ψi,0(x,μ,θ). For -1x1, -1<μ<1, Ψi,0(x,μ,θ)=f3i(x,μ,θ)-j=1J(-1)jΨi,j(x,μ,θ),0θ<π,Ψi,0(x,μ,θ)=-j=1JΨi,j(x,μ,θ),π<θ2π, where f2β(x,μ,θ)=2-δ0,jπ-11f2(x,z,μ,θ)Rj(z)1-z2dz,f3i(x,μ,θ)=2-δi,0π-11f3(x,y,μ,θ)Ti(y)1-y2dy. To determine the components Ψi,j(x,μ,θ), i=1,,I, and j=1,,J we substitute Ψ(x,μ,θ), from (2.3) into (2.1) and the boundary conditions for x=±1. Multiplying the resulting expressions by (Ti(y)/1-y2)×(Rj(z)/1-z2), and integrating over y and z we obtain I×J one-dimensional transport problems, namely, μΨi,jx(x,μ,θ)+σtΨi,j(x,μ,θ)=Gi,j(x;μ,θ)-11σs(μ,θμ,θ)Ψi,j(x,μ,θ)dθdμ, with the boundary conditions Ψi,j(-1,μ,η)=f1i,j(μ,θ), where f1i,j(μ,θ)=4π2-11Ti(y)Rj(z)(1-y2)(1-z2)f1(y,z,μ,θ)dzdy,Ψi,j(1,-μ,θ)=0, for 0<μ1, and 0θ2π. Finally, Gi,j(x;μ,θ)=Si,j(x,μ,θ)-1-μ2[cosθk=i+1IAikΨk,j(x,μ,θ)+sinθl=j+1JBjlΨi,l(x,μ,θ)], with Si,j(x,μ,θ)=4π2-11Ti(y)Rj(z)(1-y2)(1-z2)S(x,μ,θ)dzdy,Aik=2π-11ddy(Tk(y))Ti(y)1-y2dy,  Bjl=2π-11ddy(Rl(y))Rj(z)1-z2dz. Now, starting from the solution of the problem given by (2.13)–(2.17) for ΨI,J(x,μ,θ), we then solve the problems for the other components, in the decreasing order in i and j. Recall that i=I+1I=j=J+1J0. Hence, solving I×J one-dimensional problems, the angular flux Ψ(x,μ,θ) is now completely determined through (2.9).

Remark 2.2.

If we deal with different type of boundary conditions, then we consider the first components Ψi,0(x,μ,θ) and Ψ0,j(x,μ,θ) for i=1,,I and j=1,,J will satisfy one-dimensional transport problems subject to the same of boundary conditions of the original problem in the variable x.

3. Analysis

Now, we solve the first-order linear differential equation system with isotropic scattering, that is, σs(μ,ϕμ,ϕ)σs= constant. Assuming isotropic scattering, the equation (2.13) is written as μΨi,jx(x,μ,θ)+σtΨi,j(x,μ,θ)=Gi,j(x;μ,θ)σ-1102πΨi,j(x,μ,θ)dθdμ, for xΩ:={(x,y):0x1,-1y1}, μ[-1,1], and θ[0,2π].

Then, we have the following theorem that is subject to the boundary conditions (2.14).

Theorem 3.1.

Consider the integrodifferential equation (3.1) under the boundary conditions (2.14), then the function Ψi,j(x,μ,θ) satisfy the following linear system of algebraic equations: n=0NDn,mpβ¯n,i,j(p,θ)-σsn=0Npα¯n,i,j(p,θ)+σtα¯n,i,j(p,θ)=-11G¯i,j(x,μ,θ)Wne(μ)dμ+n=0NDn,mβn,i,j(0,θ),  n=0NDn,mpα¯n,i,j(p,θ)-σsn=0Npβ¯n,k(p,θ)+σtβ¯n,i,j(p,θ)=-11G¯i,j(x,μ,θ)Wno(μ)dμ+n=0NDn,mα¯n,i,j(0,θ).

Proof.

For this problem we expand the angular flux in terms of the Walsh function in the angular variable with its domain extended into the interval [-1,1]. To end this, the Walsh function Wn(μ) are extended in an even and odd fashion as follows, see : Wne(μ)={Wn(μ)if  μ0,Wn(-μ)if  μ<0,Wno(μ)={Wn(μ)if  μ0,-Wn(-μ)if  μ<0, for n=0,1,,N. The important feature of this procedure relies on the fact that a function f(μ) is defined in the interval [-1,1] it can be expanded in terms of these extended functions in the manner: f(μ)=n=0[anWne(μ)+bnWno(μ)], where the coefficients an and bn are determined as an=12-11f(μ)Wne(μ)dμ,bn=12-11f(μ)Wno(μ)dμ. So, in order to use the Walsh function for the solution of the problem (3.1), the angular flux is approximated by the truncated expansion: Ψi,j(x,μ,θ)=n=0N[αn,i,j(x,θ)Wne(μ)+βn,i,j(x,θ)Wno(μ)]. Inserting this expansion into the linear transport (3.1), it turns out n=0N[{μαn,i,jx(x,θ)+σtαn,i,j(x,θ)}Wne(μ)+{μβn,i,jx(x,θ)+σtβn,i,j(x,θ)}Wno(μ)]=n=0Nσs[-1102παn,i,j(x,θ)Wne(μ)dθdμ+-1102παn,i,j(x,θ)Wno(μ)dθdμ]+Gi,j(x,μ,θ). Multiplying (3.7) by Wme, m=0,,N and integrating over the interval [-1,1], results: n=0N[βn,i,jx(x,θ)-11μWno(μ)Wne(μ)dμ]+σtαn,i,j(x,θ)-11Wne(μ)Wme(μ)dμ=n=0Nσs[02παn,i,j(x,θ)dθ-11Wno(μ)Wno(μ)dμ]+-11Gi,j(x,μ,θ)Wne(μ)dμ. Similarly, multiplying (3.7) by Wm0, m=0,,N and integrating yields: n=0N[αn,i,jx(x,θ)-11μWno(μ)Wne(μ)dμ+σtβn,i,j(x,θ)-11Wn0(μ)Wm0(μ)dμ]=n=0Nσs[02πβn,i,j(x,θ)dθ-11Wno(μ)Wno(μ)dμ]+-11Gi,j(x,μ,θ)Wn0(μ)dμ. The integrals appearing in (3.8) and (3.9) are known and are given  as Dn,m=12-11μWno(μ)Wme(μ)dμ=01μW(n+m)mod2(μ) or Dn,m={12if  n=m,-2-(k+2),if  (n+m)mod2=2kk  natural,0at  another  case, where the notation (n+m)mod2 denotes the mod2 sum of the binary digits n and m  n=0NDn,mpβ¯n,i,j(p,θ)-σsn=0Npα¯n,i,j(p,θ)+σtα¯n,i,j(p,θ)=-11G¯i,j(x,μ,θ)Wne(μ)dμ+n=0NDn,mβn,i,j(0,θ),  n=0NDn,mpα¯n,i,j(p,θ)-σsn=0Npβ¯n,k(p,θ)+σtβ¯n,i,j(p,θ)=-11G¯i,j(x,μ,θ)Wno(μ)dμ+n=0NDn,mα¯n,i,j(0,θ).

4. Operational Tau Method and Converting Transport Equation

The operational approach to the Tau method proposed by  describes converting of a given integral, integrodifferential equation or system of these equations to a system of linear algebraic equations based on three simple matrices: γ=[0100010010:::],η=[0100200030:::],l=[010001200130:::]. We recall the following properties .

Lemma 4.1.

Let un(x) be a polynomial as un(x)=i=0naixi=i=0aixi=anX, then we have

(dr/dxr)un(x)=anηrX, r=1,2,3,,

xrun(x)=anγrX, r=1,2,3,,

axun(x)dx=anlX|ax=anlX-anlA,

where an=[a0,a1,,an,0,0,] and A=[1,a,a2,,an,]T.

Now we solve (3.12) using the Tau method, for this we consider this equation because is equivalent μΨi,jx(x,μ,θ)+σtΨi,j(x,μ,θ)=Gi,j(x;μ,θ)+-1102πσs(μ,θ)Ψi,j(x,μ,θ)dθdμ for xΩ:={(x,y):0x1,-1y1}, μ[-1,1], and θ[0,2π] subject to the following boundary conditions (2.14).

In order to convert (5.1) to a system of linear algebraic equations we define the linear differential operator D of order ζ with polynomial coefficients defined by D:=r=0ζpr(x)rxr we will write forpr(x):=k=0αrprkxk, where αr is the degree of pr(x) and pr̲=(pr0,,prαr,0,0,), Υ̲=(1,μ,μ2,)T. We notice that μ is the independent variable and will be defined in a finite interval.

4.1. Matrix Representation for the Different Parts

Let V={v0(μ),v1(μ),} be a polynomial basis given by V̲:=VΥ̲, where V is nonsingular lower triangular matrix and degree (vi(μ))i, for i=0,1,2,. Also for any matrix P, Pv=VPV-1.

Matrix Representation for <inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M182"><mml:mi>D</mml:mi><mml:msub><mml:mrow><mml:mi mathvariant="normal">Ψ</mml:mi></mml:mrow><mml:mrow><mml:mi>i</mml:mi><mml:mo>,</mml:mo><mml:mi>j</mml:mi></mml:mrow></mml:msub><mml:mo stretchy="false">(</mml:mo><mml:mi>x</mml:mi><mml:mo>,</mml:mo><mml:mi>μ</mml:mi><mml:mo>,</mml:mo><mml:mi>θ</mml:mi><mml:mo stretchy="false">)</mml:mo></mml:math></inline-formula>

Theorem 4.2 (see [<xref ref-type="bibr" rid="B13">16</xref>]).

For any linear differential operator D defined by (5.2) and any series Ψi,j(x,μ,θ):=a̲V̲,a̲̲:=(Ψ0,j(x,θ),Ψ1,j(x,θ),Ψ2,j(x,θ),,Ψi,0(x,θ),Ψi,1(x,θ),Ψi,2(x,θ),,), we have DΨi,j(x,μ,θ)=a̲̲ΞvV̲, where Ξv=Vi=0ζηipi(γ)V-1.

Matrix Representation for the Integral Form

Let us assume that σs(μ,θ)=j=0nσs(θ)vj(μ),Ψi,j(x,μ,θ)=i,j=0nΨi,j(x,θ)vi(μ)=a̲V̲, then we can write -1102πσs(μ,θ)Ψi,j(x,μ,θ)dθdμ=i,j=0n02πσs(θ)Ψi,j(x,μ,θ)dθ-11vj(μ)vi(μ)dμ=a̲KV̲, where K=[j=0nk0,jϖj0i,j=0nk0,jϖij00j=0nk0,jϖiji,j=0nk0,jϖij00], with ki,j=02πσs(θ), Ψi,j(x,μ,θ)dθ, and ϖij=-11vi(μ)vj(μ)dμ for i,j=0,1,,n.

Matrix Representation for the Boundary Conditions

Ψi,j(x,μ,θ)=i=0k=0ζΨjk(x,θ)vj(μ)=a̲Col[Ψ0,j(x,θ),Ψ1,j(x,θ),,Ψi,0(x,θ),Ψi,1(x,θ),,]=a̲Bj, where j=1,,ζ. It follows from (5.4) and (5.5) that μΨi,jx(x,μ,θ)+σtΨi,j(x,μ,θ)=-1102πσs(μ,θ)Ψi,j(x,μ,θ)dθdμ=1μ[a̲KV̲-a̲̲ΞvV̲].   Let Gi,j(x;μ,θ)=d=0nGi,jd(x;θ)vd(μ)=Gi,j̲̲(x;θ)V̲ with Gi,j=(Gi,j0,,Gi,jn,0,0,).

5. Error Estimation

Consider now the discrete ordinates (SN) approximation of the equation (2.13) for m=1,,M,μmΨα,β,Nx(x,μm,θ)+σtΨα,β,N(x,μm,θ)=n=1MωnΨα,β,N(x,μm,θ)+Gα,β(x,μm,θ). In this part, we evaluate an error estimator for the approximate solution of (2.13), we suppose that equations (2.13) and (5.1) have the same boundary conditions. Let us callϵm(x,μ,θ)=Ψi,j(x,μm,θ)-Ψi,j,m(x,μ,θ) this error function of the approximate solution Ψi,j,m to Ψi,j where Ψi,j is the exact solution of (2.13). Hence, Ψi,j(x,μm,θ) satisfies the following problem:μmΨα,β,Nx(x,μm,θ)+σtΨα,β,N(x,μm,θ)=n=1MωnΨα,β,N(x,μm,θ)+G  α,β(x,μm,θ)+HN(μm). We can evaluate the perturbation term HN(μm) by substituting the computed solution into the equationHN(μm)=μmΨα,β,Nx(x,μ,θ)+σtΨα,β,N(x,μ,θ)-n=1MωnΨα,β,N(x,μm,θ)-G  α,β(x,μm,θ) after doing some algebraic manipulations, the error functions ϵm(μ) satisfies the problemHN(μm)=μmϵm(x,μ,θ)x+σtϵm(x,μ,θ)-n=1Mωnϵm(x,μ,θ).

6. Conclusion

In general, obtaining solutions of some integrodifferential equations are usually difficult. In our recent works we have used Walsh functions, Chebyshev polynomials and Lengendre polynomials in order to reduces these kind of equations. However our present work suggests that the Tau method can be a good approximation to the exact solutions. The application of the Tau method by using the orthogonal polynomials will be considered as a future work.

Acknowledgments

The authors thank the referee(s) for the helpful and significant comments that bring the attention of authors to the references  which update the bibliography. The authors gratefully also acknowledge that this research was partially supported by the University Putra Malaysia under the ScienceFund Grant no. 06-01-04-SF1050 and Research University Grant Scheme no. 05-01-09-0720RU.