The Boundary Elements Method forMagneto-Hydrodynamic (MHD) Channel Flows at

High Hartmann Numbers

Hossein Hosseinzadeha, Mehdi Dehghana,∗, Davoud Mirzaeib

aDepartment of Applied Mathematics, Faculty of Mathematics and Computer Science,Amirkabir University of Technology, No. 424, Hafez Ave.,15914, Tehran, Iran

bDepartment of Mathematics, University of Isfahan, 81745-163 Isfahan, Iran

May 20, 2012

Abstract

In this article the constant and the continuous linear boundary elements methods(BEMs) are given to obtain the numerical solution of the coupled equations invelocity and induced magnetic field for the steady magneto-hydrodynamic (MHD)flow through a pipe of rectangular and circular sections having arbitrary conductingwalls. In recent decades, the MHD problem has been solved using some variationsof BEM for some special boundary conditions at moderate Hartmann numbers upto 300. In this paper we develop this technique for a general boundary condition(arbitrary wall conductivity) at Hartmann numbers up to 105 by applying some newideas. Numerical examples show the behavior of velocity and induced magnetic fieldacross the sections. Results are also compared with the exact values and the resultsof some other numerical methods.

Keywords: 2D boundary elements method (BEM); magneto-hydrodynamicequation (MHD); High Hartmann numbers.

AMS classification: 65N38

∗E-mail addresses:h hosseinzadeh@aut.ac.ir (H. Hosseinzadeh),mdehghan@aut.ac.ir, Corresponding author (M. Dehghan),d.mirzaei@sci.ui.ac.ir (D. Mirzaei).

1

1 Introduction

The boundary elements method (BEM) [1, 2, 3] is a suitable numerical method for solvinglinear partial differential equations which have been formulated as boundary integralequations. BEM is applicable to problems for which fundamental solutions have beencalculated [5]. The homogenous advection-diffusion (AD) equation

∇2u(x, y) + v.∇u(x, y) + c u(x, y) = 0,

is one of those equations where the corresponding fundamental solution can be derived[3, 4]. As will be completely described in this article, the coupled magneto-hydrodynamic(MHD) equations can be simplified and solved by means of change of variables and fun-damental solution of advection-diffusion equation.

The use of conducting fluids in pipes started from Williams works (1930) who per-formed experiments using a copper sulphate solution measuring a dc voltage across thechannel. Hartmann and Lazarus (1937) used mercury in their experiments instead of elec-trolytes. Shercliff [6, 7] made a detailed theoretical and experimental study for the flowin circular pipes especially because of its utility in electromagnetic flow measurements.The study of the flow of conducting fluids in the presence of transverse magnetic fieldshas attracted attention owing to its applications in such diversified fields as astrophysics,geology, power generation, flow-metry, thermonuclear reactor technology, designing cool-ing systems with liquid metals, MHD generators, accelerators, pumps, etc. The geometryof the section has been taken as a circle, rectangle, ellipse, sector, etc.

The exact solutions of steady MHD flows are available only for some simple geometriessubject to simple boundary conditions [8, 9]. Singh and Lal [10, 11, 12] obtained thenumerical solutions of MHD flows through pipes of various cross-sections using finitedifference methods (FDM) and finite element methods (FEM) for Hartmann numbersless than 10. Sezgin and Koksal [13] extended these studies to moderate Hartmannnumbers using FEM. Other related MHD studies include those of Sezgin [14], Ramos andWinowich [15], Demendy and Nagy [16], Barrett [17], Sezgin and Aydin [18], Bozkaya andSezgin [19] and Sezgin and Bozkaya [20], Verardi et al. [21, 22], and Bourantas et al. [31].The latest considered the problem at Hartmann numbers up to 105 using a collocationmeshless method.

Gupta and Singh [23] obtained the exact solutions for unsteady flows in some spe-cial cases. Singh and Lal [24] studied the FEM solution of time-dependent MHD flowequations. Seungsoo and Dulikravich [25] proposed a FDM scheme for the 3-dimensionalunsteady MHD flow together with temperature field. Sheu and Lin [26] presented aconvection-diffusion- reaction model for solving the unsteady MHD flow using a FDMscheme. The stabilized FEM for solution of the 3-dimensional time-dependent MHDequations was given by Salah et al. [27]. Bozkaya and Sezgin employed the dual reci-procity boundary element method (DRBEM) [28, 19] for the non-conducting walls andalso the time-domain BEM [29] for arbitrary wall conductivity unsteady MHD flow. In[30, 31, 32, 33] some variations of meshless methods presented for unsteady MHD prob-lem. Also in [34] a combination of finite volume and spectral elements developed for this

2

problem.Sezgin, Han and Bozkaya [18, 19] have solved the MHD problem for moderate Hart-

mann numbers up to 300 using BEM. They have also considered some special cases of theboundary conditions. However there are some valuable works in the MHD literature (see[43, 44, 45, 46]) which confirm the important range of the Hartmann number in indus-trial applications is the values between 102 and 106. In the current work we are going todevelop the BEM for the MHD problem at high Hartmann numbers up to 105 for generalboundary conditions. Up to the best of our knowledge, this is the first paper in the BEMwhich considers the MHD problems at high Hartmann numbers. To get this, three newtechniques have been presented:

• A change of variables technique,

• Calculating the singular boundary integrals in an elegant way,

• Applying a modified fundamental solution.

The next Section describes the governing equations and a suitable change of variableswhich performs everything for implementation of BEM. In Section 3 the BEM’s formula-tions for constant and linear approximations are outlined and the method of calculationof singular boundary integrals is discussed. Besides, modified fundamental solutions areperformed to deal with the high values of Hartmann number. In Section 4 some numericalresults are considered.

2 Governing equations

Suppose that Ω is the cross-section of a channel, in which a conducting fluid flows. Alsosuppose Γ is the boundary of Ω. The Cartesian coordinates (x, y, z) are chosen such thatxy-plane lies in the computational domain Ω and z be the perpendicular vector on it tothe direction along which the fluid flows. When the basic equations of fluid mechanicsand the Maxwell equations are joined to each other, the coupled system of equations ofthe velocity V (x, y) and the induced magnetic field B(x, y) are obtained. The coupledsystem is

∇2V +M · ∇B = −1,

∇2B +M · ∇V = 0,(2.1)

when the fluid is viscous and incompressible. Symbol “·” indicates the inner productoperator. Hartmann number H 6= 0 is the size of vector M = (Mx,My). In the otherwords, if α is the angle between the y-axis and the applied magnetic field of intensity B0

in the xy-plane, the constant values of Mx and My can be calculated as:

Mx = H sin(α),

My = H cos(α).

3

The general form of the boundary conditions on Γ is [37]:

V = 0 ,

∂B

∂n+ λB = 0,

(2.2)

where n is the outward normal unit vector to the boundary Γ, and λ is conductivityparameter. There are generally two special cases which are important in physical means.The first case is λ = ∞ (in non-conducting walls), and the second case is λ = 0 (for theperfectly conducting walls). For λ =∞ and λ = 0 the second boundary condition can bereduced to B = 0 and ∂B/∂n = 0, respectively [37].

Note that the second boundary condition in (2.2) can be written as:

∂B

∂n+ λ1B = 0 , on Γ1 ,

∂B

∂n+ λ2B = 0 , on Γ2 ,

where Γ = Γ1 ∪ Γ2 and Γ1 ∩ Γ2 = φ. When λ1 = ∞ and λ2 = 0, Γ1 and Γ2 are theinsulating and conducting parts of the boundary, respectively. In this case B = 0 on Γ1

and ∂B/∂n = 0 on Γ2.The governing equation (2.1) may be decoupled to

∇2W1 +M · ∇W1 = −1 ,

∇2W2 −M · ∇W2 = −1 ,(2.3)

by the change of variables W1 = V + B and W2 = V − B. In this case the boundaryconditions (2.2) will be replaced by the coupled boundary conditions

W1 = −W2 ,

∂W1

∂n+ λW1 =

∂W2

∂n+ λW2 .

(2.4)

Since H 6= 0, it is obvious that Mx 6= 0 or My 6= 0. Suppose that Mx 6= 0. If we setU1 = W1 + x/Mx and U2 = W2 − x/Mx, Eqs. (2.3) and (2.4) convert to

∇2U1 +M · ∇U1 = 0 , (2.5)

∇2U2 −M · ∇U2 = 0 , (2.6)

U1 + U2 = 0 , (2.7)

∂U1

∂n+ λU1 −

∂U2

∂n− λU2 = 2

nx

Mx

+ 2λx

Mx

, (2.8)

where n = (nx,ny), nx = ∂x/∂n and ny = ∂y/∂n.Note that when Mx = 0 and My 6= 0 the new variables U1 = W1 + y/My and U2 =

W2−y/My can be used. In this situation, (2.5)-(2.7) remain unchanged, but (2.8) changesto

∂U1

∂n+ λU1 −

∂U2

∂n− λU2 = 2

ny

My

+ 2λy

My

.

In the next section, we formulate the BEM for (2.5)-(2.8).

4

3 The boundary elements method

We apply the conventional boundary elements method to (2.5)-(2.6) and then use (2.7)-(2.8) to impose the boundary conditions. Let U∗1 and U∗2 be the fundamental solutionsof Eqs. (2.5) and (2.6), respectively. This means that, U∗1 and U∗2 satisfy the followingrelations:

∇2U∗1 −M · ∇U∗1 = −δ(x− ξ, y − η) , (3.1)

∇2U∗2 +M · ∇U∗2 = −δ(x− ξ, y − η) , (3.2)

where δ(x, y) is Dirac delta function, and (ξ, η) and (x, y) are the source and the fieldpoints, respectively. Multiplying (2.5) by U∗1 , taking integration over Ω, applying theGreen’s second identity and then using (2.5) and (3.1) yield∫

Γ

[U1∂U∗1∂n− U∗1

∂U1

∂n

]dΓ =

∫Ω

[U1∇2U∗1 − U∗1∇2U1

]dΩ

=

∫Ω

[U1

(− δ(x− ξ, y − η) +M · ∇U∗1

)+ U∗1

(M · ∇U1

)]dΩ

= −c(ξ, η)U1(ξ, η) +

∫Ω

M · ∇(U1U∗1 ) dΩ,

(3.3)

where the value of c(ξ, η) depends on the location of the source point p = (ξ, η). Gen-erally c(ξ, η) = α(ξ, η)/2π such that α(ξ, η) is the internal angle of the boundary at thecollocation point p. It is clear that α(ξ, η) = 2π when p is inside Ω, and α(ξ, η) = π whenp is located on the smooth parts of Γ. Employing the divergence theorem (pages 13-15 of[5]), Eq. (3.3) can be rewritten as

c(ξ, η)U1(ξ, η) =

∫Γ

[U∗1∂U1

∂n− U1

∂U∗1∂n

]dΓ +

∫Γ

(U1U∗1 )M · n dΓ. (3.4)

The same treatments on (2.6), give

c(ξ, η)U2(ξ, η) =

∫Γ

[U∗2∂U2

∂n− U2

∂U∗2∂n

]dΓ−

∫Γ

(U2U∗2 )M · n dΓ. (3.5)

The fundamental solutions U∗1 and U∗2 are given in [3] and [4] as:

U∗1 (ξ, η) =1

2πexp

(1

2M · r

)K0

(H2r),

U∗2 (ξ, η) =1

2πexp

(− 1

2M · r

)K0

(H2r),

(3.6)

where r = (rx, ry) is a vector obtained by subtraction of the field point (x, y) from thesource point (ξ, η), i.e. rx = x− ξ and ry = y− η, r is length of r, i.e. r =

√r2x + r2

y, andK0 is the zero order modified Bessel function of the second kind.

Substituting Eq. (3.6) to Eqs. (3.4) and (3.5), and discretizing the boundary Γ, theconstant and linear BEMs are obtained, as will be described in the forthcoming subsection.

5

3.1 Constant and Linear BEMs

Let p = (ξ, η) be a source point belongs to Γ ∪ Ω, and suppose Q1 = ∂U1/∂n andQ2 = ∂U2/∂n. If we approximate the exact boundary Γ by the jointed segment lines Γj,j = 1, ..., N and use the constant approximations for functions U1, U2, Q1 and Q2, Eqs.(3.4)–(3.6) yield

c(p)U1(p) =N∑j=1

[Q1,jG

1p,j + U1,jH

1p,j

],

c(p)U2(p) =N∑j=1

[Q2,jG

2p,j + U2,jH

2p,j

],

(3.7)

where N is the number of the boundary elements, and U1,j, U2,j, Q1,j, Q2,j are the valuesof the unknown functions U1, U2, Q1, Q2 at the mid point of the j-th boundary element,Γj. Also

(a) : G1p,j =

1

2π

∫Γj

exp(1

2M · r

)K0

(H2r)dΓj,

(b) : G2p,j =

1

2π

∫Γj

exp(− 1

2M · r

)K0

(H2r)dΓj,

(c) : H1p,j =

1

2π

∫Γj

exp(1

2M · r

)[1

2M · nK0

(H2r)

+H

2rK1

(H2r)r · n

]dΓj,

(d) : H2p,j =

1

2π

∫Γj

exp(− 1

2M · r

)[− 1

2M · nK0

(H2r)

+H

2rK1

(H2r)r · n

]dΓj,

(3.8)

where K1 is the first order modified Bessel function of the second kind.If we choose the source points p1, p2, ..., pN coincide with the mid points of the bound-

ary elements Γ1,Γ2, ...,ΓN , respectively, we get the following matrix forms

G1Q1 +H1U1 = 0, G2Q2 +H2U2 = 0, (3.9)

where

G1(i, j) = G1pi,j, G2(i, j) = G2

pi,j,

H1(i, j) = H1pi,j− 1

2δi,j, H2(i, j) = H2

pi,j− 1

2δi,j,

such that δi,j is Kronecker delta function. Finally Eqs. (2.7)-(2.8) and (3.9) lead to thesystem

H1 G1 0 00 0 H2 G2

I 0 I 0λI I −λI −I

U1

Q1

U2

Q2

=

000C

, (3.10)

6

where I (= IN×N) is the identity matrix, and C (= CN×1) is a matrix obtained from theright hand side of Eq. (2.8). The interior values of U1 and U2 can be calculated using Eq.(3.7) for p ∈ Ω, providing the linear system (3.10) is solved using some proper numericallinear algebra technique.

The constant BEM is the simplest approach. However to get more accurate results,we can apply the continuous linear BEM. Since the computational costs of the constantand the continuous linear BEMs are approximately the same, the use of the linear BEMhas been strongly recommended [2, 35].

Similar to the constant BEM, for the continuous linear BEM the boundary Γ discretizesto N boundary line elements. In this scheme the unknown functions U1, U2, Q1, Q2 on theboundary are interpolated linearly on each boundary element Γj, j = 1, 2, ..., N using twoend points pj and pj+1. If p = (ξ, η) be an arbitrary point, Eqs. (3.4)–(3.6) yield

c(p)U1(p) =N∑j=1

[Q1,j Q1,j+1]

G1,1p,j

G1,2p,j

+ [U1,j U1,j+1]

H1,1p,j

H1,2p,j

,

c(p)U2(p) =N∑j=1

[Q2,j Q2,j+1]

G2,1p,j

G2,2p,j

+ [U2,j U2,j+1]

H2,1p,j

H2,2p,j

,

(3.11)

where U1,j, U2,j, Q1,j, Q2,j are the exact values of the unknown functions at pj. Note thatin Eq. (3.11) we have Um,N+1 = Um,1 and Qm,N+1 = Qm,1 for m = 1, 2, and

(a) : G1,mp,j =

1

2π

∫Γj

Nm(t) exp(1

2M · r

)K0

(H2r)dΓj,

(b) : G2,mp,j =

1

2π

∫Γj

Nm(t) exp(− 1

2M · r

)K0

(H2r)dΓj,

(c) : H1,mp,j =

1

2π

∫Γj

Nm(t) exp(1

2M · r

)[1

2M · nK0

(H2r)

+H

2rK1

(H2r)r · n

]dΓj,

(d) : H2,mp,j =

1

2π

∫Γj

Nm(t) exp(− 1

2M · r

)[− 1

2M · nK0

(H2r)

+H

2rK1

(H2r)r · n

]dΓj,

(3.12)

where the variable t is the natural coordinate along the boundary element Γj belongs tointerval [−1, 1] taking −1 and +1 at the points pj and pj+1, respectively. And

N1(t) =1

2(1− t) , N2(t) =

1

2(1 + t) ,

are the linear shape functions. The set of algebraic equations arising from Eqs. (3.12)can be recast to the matrix form (3.9), where

G1(i, 1) = G1,1pi,1

+G1,2pi,N

, G2(i, 1) = G2,1pi,1

+G2,2pi,N

,

H1(i, 1) = H1,1pi,1

+H1,2pi,N− c(pi)δi,1 , G2(i, 1) = G2,1

pi,1+G2,2

pi,N− c(pi)δi,1 ,

7

and for j = 2, 3, ..., N ,

G1(i, j) = G1,1pi,j

+G1,2pi,j−1 , G2(i, 1) = G2,1

pi,j+G2,2

pi,j−1 ,

H1(i, j) = H1,1pi,j

+H1,2pi,j−1 − c(pi)δi,j , H2(i, 1) = H2,1

pi,j+H2,2

pi,j−1 − c(pi)δi,j .The interior unknown values can be computed using Eq. (3.11).

3.2 Calculation of Singular Boundary Integrals

To calculate the elements of matrices G1, G2, H1 and H2 we face singular boundaryintegrals. In Eqs. (3.8) and (3.12) we should integrate against the Bessel functionsK0(H/2 r) and K1(H/2 r). When r, the distance between the source point (ξ, η) andthe field point (x, y), approaches to zero, the Bessel functions tend to infinity. In thiscase the conventional numerical methods break down to produce accurate results. Toovercome this drawback we propose a new technique which is based on simplificationof the Bessel functions in an elegant way. For small values of z (i.e. z → 0) we haveK0(z) ≈ − ln(z/2) − γ and K1(z) ≈ 1/z where γ is Euler number. We define functionsF ∗1 and F ∗2 as

F ∗1 = U∗1 +1

2πln(r) =

1

2π

(1

2M · r

)K0

(H2r)

+1

2πln(r),

F ∗2 = U∗2 +1

2πln(r) =

1

2πexp

(− 1

2M · r

)K0

(H2r)

+1

2πln(r).

These functions are bounded as r tends to zero, and

limr→0

F ∗1 = − 1

2π(ln(H/4) + γ) = lim

r→0F ∗2 .

Consequently, since

U∗1 = F ∗1 −1

2πln(r), U∗2 = F ∗2 −

1

2πln(r),

the boundary integrals in Eqs. (3.8) and (3.12) can be written as subtractions of two kindsof boundary integrals including regular kernels F ∗k , k = 1, 2 and simple weak singularkernel 1/(2π) ln(r). The formers can be calculated easily by the use of the conventionalnumerical methods due to the regularity and the latest is the fundamental solution ofLaplace equation. There are many valuable techniques to deal with the singularitiesappearing in the boundary integrals for Laplace equation [38, 39, 40, 36]. Applying theabove-mentioned substitutions, equations in (3.8) are changed to

(a) : G1pj ,j

=

∫Γj

F ∗1 dΓj −Lj2π

(ln (Lj/2)− 1) ,

(b) : G2pj ,j

=

∫Γj

F ∗2 dΓj −Lj2π

(ln (Lj/2)− 1) ,

(c) : H1pj ,j

=1

2M · nG1

pj ,j,

(d) : H2pj ,j

= −1

2M · nG2

pj ,j,

8

where the boundary integrals with kernel ln(r) are calculated exactly, and Lj is thelength of j-th boundary element. For continuous linear BEM, from Eq. (3.12) we havethe following equations

(a) : G1pj ,j

=

∫Γj

Nm(t)F ∗1 dΓj − Im1 , G1pj+1,j

=

∫Γj

Nm(t)F ∗1 dΓj − Im2 ,

(b) : G2pj ,j

=

∫Γj

Nm(t)F ∗2 dΓj − Im1 , G2pj+1,j

=

∫Γj

Nm(t)F ∗2 dΓj − Im2 ,

(c) : H1pj ,j

=1

2M · nG1

pj ,j, H1

pj+1,j=

1

2M · nG1

pj+1,j,

(d) : H2pj ,j

= −1

2M · nG2

pj ,j, H2

pj+1,j= −1

2M · nG2

pj+1,j,

for m = 1, 2, such that

I11 = I2

2 =Lj4π

(ln(Lj)−

3

2

),

I21 = I1

2 =Lj4π

(ln(Lj)−

1

2

).

3.3 Modified Fundamental Solution for Large Values of Hart-mann Number

Boundary elements solution of the MHD equation fails at high values of H due to thelimitation of machine precision. In Eq. (3.6) when the multiplication Hr grows, theexponential terms in definitions of U∗1 and U∗2 approach to infinity and the Bessel termsapproach to zero. These cause roundoff errors in calculations and lead to unsatisfactorynumerical results. To overcome this drawback we can modify the fundamental solutionsU∗1 and U∗2 for large values of H, using the fact that

Kα(z) ≈√

π

2zexp(−z) , as z →∞ , (3.13)

for α = 0, 1. Regarding to (3.13) for large values of Hr (say, the values larger than 50),Eq. (3.6) can be approximated by

U∗1 (ξ, η) ≈ 1

2√πHr

exp(1

2M · r − H

2r),

U∗2 (ξ, η) ≈ 1

2√πHr

exp(− 1

2M · r − H

2r).

(3.14)

9

Thus, (3.8) can be replaced by

(a) : G1p,j =

1

2√πH

∫Γj

1√r

exp(1

2M · r − H

2r)dΓj,

(b) : G2p,j =

1

2√πH

∫Γj

1√r

exp(− 1

2M · r − H

2r)dΓj ,

(c) : H1p,j =

1

4√π

∫Γj

1√r

exp(1

2M · r − H

2r)[ 1√

HM · n +

√H

rr · n

]dΓj,

(d) : H2p,j =

1

4√π

∫Γj

1√r

exp(− 1

2M · r − H

2r)[− 1√

HM · n +

√H

rr · n

]dΓj.

Similar relations can be written for the linear BEM by substitution (3.14) into (3.12)when Hr grows up.

4 Numerical examples

As we mentioned, we assume that the fluid flows in z direction in the duct and weconsider the MHD equation in a cross-section of duct in xy-plane. Cross-sections canbe assumed to be of rectangular or circular shapes. Besides, we assume that the fluid isviscous, incompressible and electrically conductive. Functions V (x, y) and B(x, y) are thez components of the velocity and induced magnetic field, respectively.

4.1 Square cross-section

At first, a duct with a square cross-section has been considered. Authors of [41] solved thisproblem for H ≤ 300 and α = π/2 by the constant BEM in some especial cases of bound-ary conditions. We extend it for moderate and high values of H (500 ≤ H ≤ 105), fordifferent values of α , and general boundary conditions. Suppose that the computationaldomain Ω is the square |x|, |y| ≤ 1.

Let H = 20. We implement the constant BEM for N = 16 and evaluate the boundaryintegrals using a 10-point Gauss quadrature. To see the effect of increasing the con-ductivity parameter λ, Fig. 1 is presented for H = 20, α = 0 and various values of λ(λ = 0 , 1 , 10, 100 and ∞). As λ increases, the graphs along the x-axis (y = 0), convergeto the case λ = ∞ and show the behavior of solution of MHD flow with insulated walls.In [34] this problem has been solved using a spectral method. Fig. 1 confirms the resultsof [34]. Fig. 2 is performed for the purpose of demonstrating the effect of various valuesof α along the x-axis (y = 0), for λ = ∞. As can be seen from this figure, the absolutevalues of the induced magnetic field along the x-axis are reduced when α is increased from0 to π/2 as is mentioned in [34].

Now we turn to some higher values of Hartman number. For H = 500, the boundaryΓ is uniformly discretized to 40 and 32 boundary elements for constant and linear BEMs,respectively. The exact values [9] and the BEM results of the velocity have been compared

10

in Fig. 3 when α = π/2 and λ =∞. The results are presented for x ∈ [0.98, 1] (where themost actions take place) and y = 0. Note that the velocity is symmetric with respect toy-axis and it is constant for interior points far from the boundary. The maximum relativeerrors for x ∈ [−1, 1], y = 0, λ =∞ and α = π/2 are smaller than 1e− 4 for the constantand linear BEM when a 100-point Gauss quadrature scheme is employed (from here on,the number of integration points is denoted by ng). Moreover, a comparison is carried outfor ng = 10 and 100 in this figure. It is clear that the accuracy of method strongly dependson the accuracy of the quadrature formula. Taking closer look at Fig. 3, one can see thatthe higher errors occur for points very close to the boundary (for example x ∈ [0.999, 1]).The main reason is the near singularity problem which appears when the internal sourcepoint is significantly close to its corresponding boundary element. We leave this for afuture work and now refer the interested readers to [38, 36] for more details.

In Table 1 the BEM solutions are compared with the exact solutions of [9], the numer-ical solutions obtained by the finite element method in [42], and the results of meshlesspoint collocation method (MPCM) [31] for H = 500 at some grid points.

Table 1The velocity and the induced magnetic field at H = 500

x y VFEM VMPCM VBEM Vexact BFEM BMPCM BBEM Bexact

0.00 0.00 200e− 5 200e− 5 200e− 5 200e− 5 0.00000 0.00000 0.00000 0.000000.25 0.25 200e− 5 200e− 5 200e− 5 200e− 5 −500e− 6 −500e− 6 −500e− 6 −500e− 6

0.50 0.50 200e− 5 200e− 5 200e− 5 200e− 5 −100e− 5 −100e− 5 −100e− 5 −100e− 5

0.75 0.75 200e− 5 200e− 5 200e− 5 200e− 5 −150e− 5 −150e− 5 −150e− 5 −150e− 5

0.00 0.75 200e− 5 200e− 5 200e− 5 200e− 5 0.00000 0.00000 0.00000 0.00000

0.25 0.75 200e− 5 200e− 5 200e− 5 200e− 5 −100e− 5 −100e− 5 −100e− 5 −100e− 5

0.25 0.50 200e− 5 200e− 5 200e− 5 200e− 5 −500e− 6 −500e− 6 −500e− 6 −500e− 6

The velocity V and the magnetic field B are shown in Fig. 4 for constant BEM. In thisfigure H = 500, α = π/4, π/3, π/2, λ = ∞ and the boundary has been discretized to 40boundary elements. The same results for α = π/2 were achieved in [31] where a meshfreemethod was applied using 1000 meshless nodes. Also the authors of [42] used 6400 nodesto solve a similar problem. In this paper we use only 40 boundary nodes leading to anabsolutely cheaper numerical scheme. As it can be seen from Fig. 4, when α = π/2 thevelocity is symmetric with respect to x-axis and y-axis. Similarly, when α = π/4 it issymmetric with respect to lines y = ±x. The induced magnetic field is symmetric withrespect to y-axis and line y = x for α = π/2 and α = π/4, respectively. Moreover it isanti-symmetric with respect to the x-axis and line y = −x for α = π/2 and α = π/4,respectively.

For H = 1000 some similar comparisons are done and shown in Fig. 5. The results areprovided for x ∈ [0.99, 1], y = 0, α = π/2 and λ =∞. The numbers of boundary elementsare 80 and 64 for the constant and linear BEMs, respectively. When the Hartmann number

11

increases, the main action takes place near the boundary. In this case the maximumrelative error is 1e − 4 for the constant and linear BEMs. The near singularity problemis evident when x ∈ [0.9995, 1] and ng = 10.

The velocity and the induced magnetic field for H = 1000 are depicted in Fig. 6by contour lines for λ = ∞. The number of boundary elements is chosen to be 80 forthe constant BEM. The results are in good agreement with those given in [31, 42]. Forα = π/2, maximum and minimum values of the velocity are 1.000e − 3 and 1e − 15 (≈machine’s epsilon in double precession), respectively. They are 9.921e−4 and −9.921e−4for the induced magnetic field.

The results of velocity for H = 104 and H = 105 with a comparison to the exactvalues are depicted in Fig. 7 for α = π/2 and λ =∞. When H = 104 and ng = 100, themaximum relative error is 1e− 3 using N = 800 and N = 480 for the constant and linearBEMs, respectively. For H = 105 it is again 1e−3 using N = 1600 and ng = 500 for linearBEM. Note that, for higher values of H the costs of the constant BEM increase due to thenumber of boundary elements and number of integration points which are required to getaccurate results. Unfortunately, for H = 105 the results are not accurate for ng < 500.The numerical results presented in Fig. 7–(a) are not accurate for x ∈ [0.9999, 1] due tothe near singularity problem. The velocity and induced magnetic field for H = 105 areshown in Fig. 8. In this case the linear BEM is applied for λ =∞ and N = 1600.

4.2 Circle cross-section

Let the computational domain be a circle with center (0, 0) and radius 1. Here we supposeα = π/2 and λ = ∞. The contour lines of the velocity and the induced magnetic fieldfor H = 500 and H = 1000 are shown in Fig. 9. The problem has been solved using theconstant BEM where N = 40 and N = 80 for H = 500 and H = 1000, respectively. Inboth cases ng = 100.

5 Conclusion and Outlook

In the boundary elements method’s (BEM’s) literatures there are some studies on Magneto-hydrodynamic (MHD) equations. This problem has been solved for Hartmann numbersup to 300 using BEM until now. In this work we extend the BEM for Hartmann numbersup to 105. To do this, some new improvements are applied. The ideas presented in thiswork can also be applied to the other PDEs with advection and diffusion terms. Finallyit is worth pointing out that the boundary integrals are not calculated accurately whenthe number of Gaussian integration points is less than 100. A new cheaper scheme shouldbe designed to evaluate the boundary integrals in future studies. In addition, we face thenear singularity problem when we try to calculate the values of the unknown functionsat points very close to the boundary (for example r < Lj/100). In this situation theconventional numerical integration schemes (i.e. Gauss quadrature rules) cannot evaluatethe boundary integrals accurately. Therefore, to get more accurate results we have to

12

treat the near singularity problem [36].

Acknowledgments

The authors thank the anonymous referees for the constructive comments and suggestions.The third author was partially supported by the Center of Excellence for Mathematics,

University of Shahrekord.

References

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16

−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 10

0.01

0.02

0.03

0.04

0.05

λ=∞λ=100

λ=10

λ=1

λ=0

−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1−0.05

−0.04

−0.03

−0.02

−0.01

0

0.01

0.02

0.03

0.04

0.05

λ=∞λ=100

λ=10

λ=1

λ=0

Figure 1: The velocity and induced magnetic field for H = 20, α = 0, N = 16, y = 0 anddifferent values of λ.

17

−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 10

0.01

0.02

0.03

0.04

0.05

0.06

α=π/2

α=π/3

α=π/4

α=π/6

α=0

−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1−0.04

−0.03

−0.02

−0.01

0

0.01

0.02

0.03

0.04

α=pi/2

α=π/3

α=π/4

α=π/6

α=0

Figure 2: The velocity and induced magnetic field for H = 20, λ = ∞, N = 16, y = 0 anddifferent values of α.

18

0.98 0.985 0.99 0.995 10

0.5

1

1.5

2

x 10−3

exact N=40 − Con. BEM − n

g=100

N=32 − Lin. BEM − ng=100

N=40 − Con. BEM − ng=10

N=32 − Lin. BEM − ng=10

Figure 3: The velocity for H = 500, α = π/2 and λ =∞.

19

−0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

−4

−2

0

2

4

6

8

10

12

14

16x 10−4

(a) −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

−1.5

−1

−0.5

0

0.5

1

1.5x 10−3

−0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

0

5

10

15

20x 10−4

(b) −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

−2

−1.5

−1

−0.5

0

0.5

1

1.5

2x 10−3

−0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

0

0.5

1

1.5

2

x 10−3

(c) −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

−2

−1.5

−1

−0.5

0

0.5

1

1.5

2

x 10−3

Figure 4: Contours of velocity and induced magnetic field for H = 500, λ = ∞ (a) α = π/2,(b) α = π/3, (c) α = π/4.

20

0.99 0.992 0.994 0.996 0.998 10

0.2

0.4

0.6

0.8

1

x 10−3

exact N=80 − Con. BEM − n

g=100

N=64 − Lin. BEM − ng=100

N=80 − Con. BEM − ng=10

N=64 − Lin. BEM − ng=10

Figure 5: The velocity for M = 1000, α = π/2 and λ =∞.

21

−0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

0

1

2

3

4

5

6

7

8x 10−4

(a) −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

−8

−6

−4

−2

0

2

4

6

8x 10−4

−0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

2

3

4

5

6

7

8

9

10x 10−4

(b) −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1x 10−3

−0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

2

3

4

5

6

7

8

9

10

11

12x 10−4

(c) −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

−1

−0.5

0

0.5

1

x 10−3

Figure 6: Contours of velocity and induced magnetic field for M = 1000, λ =∞, (a) α = π/2,(b) α = π/3, (c) α = π/4.

22

0.999 0.9992 0.9994 0.9996 0.9998 10

0.2

0.4

0.6

0.8

1

x 10−4

exact N=800 − Con. BEM − n

g=100

N=480 − Lin. BEM − ng=100

N=800 − Con. BEM − ng=10

N=480 − Lin. BEM − ng=10

(a)

0.9998 0.9999 10

0.2

0.4

0.6

0.8

1

x 10−5

exact N=1600 − Lin. BEM − n

g=500

N=1600 − Lin. BEM − ng=70

N=1600 − Lin. BEM − ng=10

(b)

Figure 7: The velocity for (a): H = 104 and (b): H = 105 when α = π/2 and λ =∞.

23

−0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

0

1

2

3

4

5

6

7

8x 10−6

(a) −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1x 10−5

−0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

2

4

6

8

10

12x 10−6

(b) −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

−1

−0.5

0

0.5

1

x 10−5

−0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

0

2

4

6

8

10

12

14x 10−6

(c) −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

−1

−0.5

0

0.5

1

x 10−5

Figure 8: Contours of velocity and induced magnetic field for H = 105, λ =∞, (a) α = π/2 ,(b) α = π/3 , (c) α = π/4.

24

−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8x 10

−3

(a)−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1

−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

−1.5

−1

−0.5

0

0.5

1

1.5x 10

−3

−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

1

2

3

4

5

6

7

8

9x 10

−4

(b) −1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

−8

−6

−4

−2

0

2

4

6

8x 10

−4

Figure 9: Contours of velocity and induced magnetic field for (a) H = 500 (b) H = 1000 whenα = 0 and λ =∞.

25