Numerical Methods Book.pdf

Boundary Element Method for Impedance and Optical TomographyJan Sikora

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Warsaw 2007

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Contents

1 Twodimensional potential problems

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1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 1.2 Laplace equation . . . . . . . . . . . . . . . . . . . . . . . . . 21 1.2.1 Division of the boundary into constant boundary elements . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 Numerical integration of the kernels . . . . . . . . . . . 25 Division of the boundary into linear boundary elements 30 Numerical integration of the kernels . . . . . . . . . . . 33 Division of the boundary into quadratic boundary elements . . . . . . . . . . . . . . . . . . . . . . . . . . . 38 Numerical integration of the kernels . . . . . . . . . . . 41 Numerical integration of the c(r) coecient . . . . . . 47 Internal points calculation . . . . . . . . . . . . . . . . 48 Symbolic calculation . . . . . . . . . . . . . . . . . . . 51

1.2.2 1.2.3 1.2.4 1.2.5

1.2.6 1.2.7 1.2.8 1.2.9

1.3 Diusion equation . . . . . . . . . . . . . . . . . . . . . . . . . 53 3

4 1.3.1 1.3.2 1.3.3

CONTENTS Treatment of singularity . . . . . . . . . . . . . . . . . 55 Internal points calculation . . . . . . . . . . . . . . . . 56 Example . . . . . . . . . . . . . . . . . . . . . . . . . . 57

1.4 Diusion equation in frequency domain . . . . . . . . . . . . . 57 1.4.1 1.4.2 Treatment of singularity . . . . . . . . . . . . . . . . . 61 Internal points calculation . . . . . . . . . . . . . . . . 61

1.5 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62 1.5.1 1.5.2 1.5.3 Cartesian coordinate system . . . . . . . . . . . . . . . 62 Polar coordinate system . . . . . . . . . . . . . . . . . 64 Distributed source for a diusion model for light transport . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65 Point source located on the boundary for a diusion model for light transport . . . . . . . . . . . . . . . . 70 Point source located inside the region for a diusion model for light transport . . . . . . . . . . . . . . . . 76 Comparison FEM and BEM results of calculation . . . 80 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . 81

1.5.4

1.5.5

1.5.6 1.5.7

1.6 Anisotropic medium . . . . . . . . . . . . . . . . . . . . . . . 81 1.6.1 1.6.2 1.6.3 Anisotropy model . . . . . . . . . . . . . . . . . . . . . 82 Treatment of singularity . . . . . . . . . . . . . . . . . 84 Comparison FEM and BEM results . . . . . . . . . . . 86

CONTENTS

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1.7 Galerkin formulation of boundary integral equations . . . . . . 88 1.7.1 1.7.2 Analytical integrations of coincident integrands . . . . 91 Numerical integrations of coincident integrands . . . . 93

2 Three-dimensional potential problems

97

2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97 2.2 Singular and nearly singular integrals . . . . . . . . . . . . . . 99 2.3 Governing equations . . . . . . . . . . . . . . . . . . . . . . . 101 2.4 Zeroorder interpolation functions . . . . . . . . . . . . . . . . 103 2.4.1 2.4.2 2.4.3 Jacobian . . . . . . . . . . . . . . . . . . . . . . . . . . 104 Integration of nonsingular integrals over the triangle . 106 Integration of singular integrals . . . . . . . . . . . . . 107 . . . . . . . . . . . . . . . 108

2.5 Firstorder interpolation functions

2.6 Secondorder interpolation functions . . . . . . . . . . . . . . 112 2.6.1 2.6.2 2.6.3 2.6.4 Triangular boundary elements . . . . . . . . . . . . . . 112 Numerical integration of singular integrals . . . . . . . 114 Quadrilateral boundary elements . . . . . . . . . . . . 120 Integration of nonsingular integrals over the square . . . . . . . . . . . . . . . . . . . . . . 121 Integration of singular integrals over the square . . . . 122

2.6.5

2.7 Treatment of Boundary Conditions . . . . . . . . . . . . . . . 130

6 2.7.1 2.7.2 2.7.3 2.7.4

CONTENTS Dirichlet boundary conditions . . . . . . . . . . . . . . 130 Neumann boundary conditions . . . . . . . . . . . . . . 130 Robin boundary conditions . . . . . . . . . . . . . . . . 131 Mixed boundary conditions . . . . . . . . . . . . . . . 131

2.8 Nonhomogeneity . . . . . . . . . . . . . . . . . . . . . . . . . 131 2.9 Index mismatched diusive/diusive interfaces . . . . . . . . . 133 2.9.1 2.9.2 Approximate interface conditions . . . . . . . . . . . . 135 Complete interface conditions . . . . . . . . . . . . . . 140

2.10 Numerical examples . . . . . . . . . . . . . . . . . . . . . . . . 142 2.10.1 Cube . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143 2.10.2 Two concentric spheres . . . . . . . . . . . . . . . . . . 143 2.10.3 The proximity eect . . . . . . . . . . . . . . . . . . . 157 2.10.4 Results for spatially nonhomogeneous region in 2D space . . . . . . . . . . . . . . . . . . . . . . . . . . . . 158

3 Diusion model for light transport in the frequency domain

161

3.1 Governing equations . . . . . . . . . . . . . . . . . . . . . . . 162 3.1.1 3.1.2 3.1.3 Two dimensional space . . . . . . . . . . . . . . . . . . 163 Three dimensional space . . . . . . . . . . . . . . . . . 163 The Boundary Element Method . . . . . . . . . . . . . 163

CONTENTS

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3.2 Numerical Implementation . . . . . . . . . . . . . . . . . . . . 164 3.2.1 3.2.2 Jacobian . . . . . . . . . . . . . . . . . . . . . . . . . . 166 Matrix Assembly . . . . . . . . . . . . . . . . . . . . . 166

3.3 Numerical integration of singular integrals in 3D . . . . . . . . 168 3.3.1 3.3.2 3.3.3 Mapping formula for triangular constant element . . . 168 Isoparametric triangular quadratic element . . . . . . . 171 Isoparametric quadrilateral quadratic element . . . . . 175

3.4 Results for 3D . . . . . . . . . . . . . . . . . . . . . . . . . . . 177 3.4.1 3.4.2 3.4.3 Validation of numerical results . . . . . . . . . . . . . . 179 Measures of the accuracy . . . . . . . . . . . . . . . . . 179 Quadratic meshing . . . . . . . . . . . . . . . . . . . . 181

3.5 Multilayered model of the neonatal head . . . . . . . . . . . . 182

4 Light propagation in diusive media with nonscattering regions 187 4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 187 4.2 Governing equations for nonscattering sphere embedded in a diusive spherical region . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 188 4.2.1 4.2.2 The Boundary Element Method . . . . . . . . . . . . . 189 Matrix form of integral equations . . . . . . . . . . . . 193

4.3 The Form Factor . . . . . . . . . . . . . . . . . . . . . . . . . 194

8 4.3.1 4.3.2

CONTENTS The Form Factor calculated analytically . . . . . . . . 194 The Form Factor calculated numerically . . . . . . . . 195

4.4 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 196 4.4.1 4.4.2 The steady state . . . . . . . . . . . . . . . . . . . . . 196 The frequency domain solution 100MHz . . . . . . . 198

4.5 Nonscattering gap between two diusive regions of a spherical shape . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 198 4.5.1 4.5.2 4.5.3 4.5.4 4.5.5 4.5.6 Form Factor calculated analytically . . . . . . . . . . . 199 Visibility function calculated analytically . . . . . . . . 200 Visibility function calculated numerically . . . . . . . . 202 Point in triangle test . . . . . . . . . . . . . . . . . . . 204 Integral equations . . . . . . . . . . . . . . . . . . . . . 206 Matrix form of integral equations . . . . . . . . . . . . 207

4.6 Results for the void gap . . . . . . . . . . . . . . . . . . . . . 208 4.6.1 4.6.2 The steady state . . . . . . . . . . . . . . . . . . . . . 208 The frequency domain solution 100MHz . . . . . . . 208

4.7 Consistency checks for BEM solutions . . . . . . . . . . . . . . 209 4.7.1 4.7.2 Diusion equation case . . . . . . . . . . . . . . . . . . 210 RadiosityDiusion equation case . . . . . . . . . . . . 212 . . . . . . . . . . . . . . . . . . . . 213

4.8 Nuuttis 2D test example

CONTENTS 4.8.1 4.8.2 4.8.3 4.8.4

9 Analytical solution for diusive boundary conditions . 214 Analytical solution for P1 boundary conditions . . . . 221 Numerical results . . . . . . . . . . . . . . . . . . . . . 223 Multilayered neonatal head model with the CSF layer . 224

4.9 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 228

5 BEM formulation for thin layers

229

5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 229 5.2 Standard BEM formulation . . . . . . . . . . . . . . . . . . . 230 5.3 Modication for closely spaced surfaces . . . . . . . . . . . . . 232 5.4 Integration of singular integrals . . . . . . . . . . . . . . . . . 235 5.5 Future work . . . . . . . . . . . . . . . . . . . . . . . . . . . . 237 5.6 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 237

6 Wavelet based techniques for CPU time reduction

239

6.1 Discrete wavelet transform . . . . . . . . . . . . . . . . . . . . 242 6.2 Time acceleration . . . . . . . . . . . . . . . . . . . . . . . . . 244 6.3 DWT with permutations . . . . . . . . . . . . . . . . . . . . . 245 6.4 DWT when the size of the coecient matrix is not a number equal to 2n . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 247 6.5 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 249

10 7 FEMBEM coupling

CONTENTS 251

7.1 2D space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 252 7.1.1 Incorporating the BE equations to the FE ones . . . . 253

7.2 Numerical examples . . . . . . . . . . . . . . . . . . . . . . . . 257 7.2.1 7.2.2 7.2.3 Simple benchmark problem . . . . . . . . . . . . . . . 257 Two squares one immersed in the other . . . . . . . . . 262 Concentric circles . . . . . . . . . . . . . . . . . . . . . 268

7.3 2D void comparison . . . . . . . . . . . . . . . . . . . . . . . . 274 7.3.1 7.3.2 7.3.3 Boundary conditions at diusive/nondiusive interfaces275 P1 boundary conditions . . . . . . . . . . . . . . . . . 276 Diusive boundary conditions . . . . . . . . . . . . . . 277

8 Miscellaneous

279

8.1 Introduction to FEM mixed formulation . . . . . . . . . . . . 279 8.2 Mixed formulation for Laplace equation . . . . . . . . . . . . . 279 8.2.1 8.2.2 Discretization of mixed forms . . . . . . . . . . . . . . 280 Robin boundary conditions . . . . . . . . . . . . . . . . 281

8.3 Mixed formulation for Diusion equation . . . . . . . . . . . . 284 8.3.1 Three nodes triangle with three degrees of freedom P1 triangle . . . . . . . . . . . . . . . . . . . . . . . . 285 Three nodes triangle with the middle sides nodes . . . 286

8.3.2

CONTENTS 8.3.3

11 Three nodes triangle with a bubble function at the barycenter P1 + triangle . . . . . . . . . . . . . . . . . 287

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CONTENTS

List of symbols represents any potential function (electric potential [V] in Electrostatics, temperature [0 C] in heat conduction problems or photon density in Diusive Optical Tomography. T temperature [0 C] or [0 K].r) q normal derivative of ( ( ). n

G the Green function. r position vector. n unit outward vector normal to the boundary. x, y, z cartesian coordinate system. r, , z or , , z cylindrical coordinate system. domain under consideration. or S surface of the domain . or local coordinate system. Ni or Mi shape functions of the i th node. J ( ) Jacobian of transformation. [A] and [B] matrices containing integrals of the Green function normal derivative and Green function respectively. 13

14 R distance between load and observation points.

CONTENTS

wi weight function (for example in the GaussLegendre quadrature). L length of the boundary element. c(r) coecient of integral equation. k wave number [cm1 ]. D diusion coecient [cm]. K0 modied Bessel function of the second kind and zero order. K1 modied Bessel function of the second kind and rst order. L[] Laplace transform. i= 1.

H vector of magnetic eld intensity [A/m]. J vector of current density [A/m2 ]. the angular frequency [ s1 ]. f frequency [1/s]. material conductivity [1/( m)]. magnetic permeability [H/m]. In modied Bessel function of the rst kind and the n th order. A coecient depending on refractive index. c speed of light [mm/ps]. a or absorbing coecient [mm1 ]. s scattering coecient [mm1 ]. s reduced scattering coecient [mm1 ].

CONTENTS Ld diusion length [mm]. Dirac delta function. ni or i refracting index of the i th diusive region. Cn refracting index mismatch. G radiocity kernel. V Boolean visibility function. g Form Factor [1/m2 ]. Uco visibility cuto angle. s directional vector. W onedimensional wavelet transform. threshold (truncation level). F functional. coecient equal to 2 or to 3 depending on the space dimensionality.

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CONTENTS

IntroductionThere is a lot of excellent books devoted the Boundary Element Method (BEM). That is why I would not dare to write one more. The main goal of this book is only to show the problems of modelling the forward problems of Diusive Optical Tomography (DOT) or Electrical Impedance Tomography (EIT) when the Boundary Element is used in 2D or 3D space. The modern Inverse Problems are solved by the iterative process in which we have to solve many times the forward problems for which the design parameters are improved due to the Sensitivity Analysis (SA). The scope of this book is limited to the forward problems and methods presentation of their eective solution in 2D and 3D space with the aid of the Boundary Element Method. The special attention was paid to the basic of the BEM because majority of the monographes devoted this method, do not pay enough attention to the problems of singular integrals integrations. This problem is already solved by many famous authors ( maybe with the exception of the 3D Galerkin approach) but is spread in the literature, in most cases, not easily accessible. That is why it seems to be reasonable to present this problem in full details in this work. A lot of attention was devoted to the problem of modelling of the infant head for the DOT. The BEM seems to be more elastic tool than the FEM. This

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18 work tries to prove this opinion.

CONTENTS

However, some disadvantages of BEM exist, for example anisotropy of neonatal brain, which force us to develop hybrid method. Such problems are presented in the last chapters of this book. Taking above remarks into account someone may say that this book is not only addressed to the community dealing with DOT, but also to the engineers, last years students or PhD students dealing with medical or industrial tomography. This book might be helpful to the students of Technological Universities, dealing with numerical simulation used in dierent areas of technology. As an Author I am conscious that not all problems of the BEM were exhaust in this book. Many of them were merely announced. The subject is so fascinated that further research will allow to remove those disadvantages. For all critical remarks send by e-mail (the actual address may be nd in the internet) I will be extremely grateful and I will consider them carefully. Jan Sikora

Chapter 1 Twodimensional potential problems

1.1

Introduction

Starting to deal with Boundary Element (BE) method someone would compare it to its main rival, the Finite Element (FE) method. The BEM advantages and disadvantages either in 2D or 3D space can be summarized as follows. Advantages of the BE Method 1. Less data preparation time. This is a direct result of the surfaceonly modelling. Some of the authors say that it is reduction of dimensionality by one [14]. Furthermore, subsequent changes in meshes are made easier. This advantage is particularly important in problems where remeshing is required, such as inverse problems, for example optimal shape design. 2. High resolution regarding the gradient of the state function. Solution is exact and fully continuous inside the domain. This makes the BE me19

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CHAPTER 1. TWODIMENSIONAL POTENTIAL PROBLEMS thod very suitable for modelling problems of rapidly changing gradients of the state function. 3. Both state function and its normal derivative racy. n

of the same accu-

4. Less computer time and storage. For the same level of accuracy, the BE method uses less number of nodes and elements (but contrary to the FEM fully populated matrix of coecients). However, there are problems where either FEM/BEM can give rise to the smaller system and quickest solutionit depends partly on the volume to surface ratio. 5. Less unwanted information. Since internal points in BE solution are optional, the user can focus on a particular interior region rather than the whole domain. 6. Open boundary problems more easy to handle. For problems involving innite or semiinnite domains, BEM is to be favoured. Disadvantages of the BE Method 1. Unfamiliar mathematics. Integrals are more dicult to evaluate and some contain integrands which become singular. The singular integrals have a signicant eect on the accuracy of the solution, so these integrals need to be evaluated accurately. However many FE numerical procedures are directly applicable to BE solutions. 2. A fundamental solution must be found (or at least an approximate one) before BEM can be applied. 3. The interior has to be discretized in many problems (Poisson equation or nonlinear material problems). That means that the method is loosing its main advantage of the reduction dimensionality. 4. Poor behavior for thin regions (for example the void problems). 5. Fully populated nonsymmetric, not positively dened matrix and not as good conditioned as in FEM.

1.2. LAPLACE EQUATION

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So, which method is better? To answer this question we have to know what kind of problem we would like to solve. Certainly the BEM is very suitable and more accurate for the linear 2D and 3D problems with rapidly changing variables such as for example light transport ones.

1.2

Laplace equation

The dierential equation representing Laplaces equation in two dimensions can be written as follows: 2 (r) = 2 (r) 2 (r) + x2 y 2 (1.1)

where (r) represents any potential function (electric potential in electrostatics, or temperature in heat conduction problems) and x and y are the Cartesian coordinates. Consider now the physical domain of the problem.n

enclosed solution domain

n

y surface 0 x z

Figure 1.1: Region in local coordinate system

Figure 1.1 shows an arbitrary domain where the solution is sought. In this solution domain, assume that there is an interior point r (usually called the load point) of coordinates X and Y , and consider any point on the boundary r (usually called the eld point) with coordinates x and y . The use

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CHAPTER 1. TWODIMENSIONAL POTENTIAL PROBLEMS

of capital letters for the coordinates indicates a xed point, whereas small letters indicate a variable point.a)

111111111 000000000 000000000 111111111 000000000 111111111 000000000 111111111 domain 000000000 111111111 000000000 111111111 000000000 111111111

n

11111111111111111 00000000000000000 00000000000000000 11111111111111111 00000000000000000 11111111111111111 b) 00000000000000000 11111111111111111 domain 00000000000000000 11111111111111111 00000000000000000 11111111111111111 00000000000000000 11111111111111111 00000000000000000 11111111111111111 00000000000000000 11111111111111111 hole n 00000000000000000 11111111111111111 00000000000000000 11111111111111111 00000000000000000 11111111111111111 00000000000000000 11111111111111111 00000000000000000 11111111111111111 00000000000000000 11111111111111111 00000000000000000 11111111111111111 00000000000000000 11111111111111111

Figure 1.2: a) Anticlockwise numbering direction for a closed domain b) Clockwise numbering direction for an opened domain

1.2.1

Division of the boundary into constant boundary elements

The boundary of the region under consideration will be divided into M straight line segments (boundary elements) as it is shown in Fig. 1.3. The nodes where the unknown values are considered, are taken to be in the middle of each boundary element. The values of or q = are assumed to be conn stant on each element and equal to the value at the mid-node of the element. Let consider the following integral equation [3, 14]: c(r)(r) +

G(|r r |) (r )d(r ) = n

(r ) d(r ) (1.2) G(|r r |) n

where r usually called the load point and the point on the boundary, r is called eld point. The coecient c(r) will be described later in section 1.2.7 and G is the fundamental solution of Laplace equation. Now the Boundary Integral Equation (BIE), equivalent to Laplace equation, (see Eq.1.1) can be written in terms of local coordinate (see Fig. 1.4), instead of the boundary curve .


1111111111111111111111 0000000000000000000000 0000000000000000000000 1111111111111111111111 0000000000000000000000 1111111111111111111111 0000000000000000000000 1111111111111111111111 q 0000000000000000000000 1111111111111111111111 0000000000000000 1111111111111111 4 0000000000000000000000 1111111111111111111111 0000000000000000 1111111111111111 0000000000000000000000 1111111111111111111111 0000000000000000 1111111111111111 q3 0000000000000000000000 1111111111111111111111 0000000000000000 00000000000000000000001111111111111111 1111111111111111111111 0000000000000000 1111111111111111 0000000000000000000000 1111111111111111111111 i=4 0000000000000000 1111111111111111 0000000000000000000000 1111111111111111111111 0000000000000000 1111111111111111 0000000000000000000000 1111111111111111111111 0000000000000000 1111111111111111 4 j=3 0000000000000000000000 1111111111111111111111 0000000000000000 1111111111111111 3 0000000000000000000000 1111111111111111111111 0000000000000000 1111111111111111 j=4 0000000000000000000000 1111111111111111111111 0000000000000000 1111111111111111 0000000000000000000000 1111111111111111111111 i=3 0000000000000000000000 1111111111111111111111 0000000000000000000000 1111111111111111111111 0000000000000000000000 1111111111111111111111 i=5j=5 domain j=2

23

i=0

j=0

j=1

i=2

i=1

Figure 1.3: Boundary discretization for a constant elements with unknowns and q = over some part of the boundary n

11111111111111111111111 00000000000000000000000 00000000000000000000000 11111111111111111111111 00000000000000000000000 11111111111111111111111 00000000000000000000000 11111111111111111111111 q0 00000000000000000000000 11111111111111111111111 00000000000000000000000 11111111111111111111111 k=1 00000000000000000000000 11111111111111111111111 N0=1 00000000000000000000000 11111111111111111111111 00000000000000000000000 11111111111111111111111 00000000000000000000000 11111111111111111111111 =+1 00000000000000000000000 11111111111111111111111 00000000000000000000000 11111111111111111111111 00000000000000000000000 11111111111111111111111 00000000000000000000000 11111111111111111111111 00000000000000000000000 11111111111111111111111 =0 00000000000000000000000 11111111111111111111111 0 00000000000000000000000 11111111111111111111111 00000000000000000000000 11111111111111111111111 j=0 00000000000000000000000 11111111111111111111111 00000000000000000000000 11111111111111111111111 00000000000000000000000 11111111111111111111111 k=0=1

Figure 1.4: Constant element (the shape function N0 for this element is equal to one), in local coordinates system

24


It can be easily veried that the coordinates of a point on element e with the intrinsic coordinate are given: x( ) = x i + xj xj x i + 2 2 (1.3) y ( ) = yi + yj yj yi + 2 2

Substituting the global numbering of the nodes by the local system and rearranging the Eq.(1.3) we will get: 1 1 x( ) = (1 )x0 + (1 + )x1 2 2 (1.4) 1 1 y ( ) = (1 )y0 + (1 + )y1 2 2 From now on we will work in the local numbering system and use the global numbering system to extract coordinates. Now Eq.(1.4) can be written in matrix notation. x = [x, y ] =T k=1 k=0

Nk ( )xk

(1.5)

where Nk ( ) is basis interpolation function at node k th often called the shape function, x is a vector containing coordinates of a point on particular element and xk is a vector of coordinates of the k th node of that element. Note that vector x can have up to three components (x, y, z ). For the two node constant element just derived the shape functions are: 1 N = [Nk=0 ( ), Nk=1 ( )]T = (1 + k ) 2 where: is the local intrinsic coordinate (consult Fig. 1.4), and for: k=0 k = 1 (1.7) k=1 k = +1 (1.6)

The shape function is equal to one at its own node and equal to zero at the other nodes (please consult the Fig. 1.7 and Fig. 1.11).


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1.2.2

Numerical integration of the kernels

Now we turn our attention to the numerical integration of the kernels. It is important to note that the choice of the local coordinate system from -1 to +1 was not arbitrary, because it happens to be the same as the limits used in the Gaussian quadrature technique. The boundary curve is now divided into elements j and the numerical integration performed over each element using the local intrinsic coordinate rather than the boundary segment j . f (x, y )d =j 1

+1

f (x( ), y ( ))J ( )d

(1.8)

where: f means any function. The Jacobian1 of transformation is as follows: d J ( ) = = d ( dx( ) d )2 + ( dy ( ) d )2 (1.9)

In the boundary element method it will be necessary to work out the direction normal to a line or surface element. The best way to determine these directions is use of vector algebra [15]. Therefore the components of unit outward normal are given by [14]: [ ] 1 dy ( ) nx = J ( ) d where: dx( ) dN0 ( ) dN1 ( ) = x0 + x1 d d d (1.11) dy ( ) dN0 ( ) dN1 ( ) = y0 + y1 d d dThe Jacobian is shorthand for either the Jacobian matrix or its determinant, the Jacobian determinant. In most cases Jacobian means determinant.1

[ ] dx( ) 1 ny = J ( ) d

(1.10)

26


and the dierentials of the shape functions are easily determined as follows: ( ) dN0 ( ) d 1 1 = (1 ) = d d 2 2 (1.12) ( ) dN1 ( ) d 1 1 = (1 + ) = + d d 2 2 Using results of Eq.(1.12) from Eq.(1.11) we will get: dx( ) x1 x0 = d 2 (1.13) dy ( ) y1 y0 = d 2 Introducing above relations into Eq.(1.9) nally we have: d J ( ) = = d ( x1 x0 2 )2 + ( y1 y0 2 )2 1 = L 2

(1.14)

where L denotes the length of the boundary element. Taking into account above relations the expressions for components of the unit outward normal vector in case of the constant element become very simple and constant over the element. nx = y1 y0 L ny = x1 x0 L (1.15)

Now the BIE of Eq.(1.2) can be written in terms of local coordinate instead of the boundary curve , as follows: c(r)i (r) +M 1 j =0 M 1 j =0

j (r ) j (r ) n

+1 1

G(|r r |) J ( )d = n G(|r r |)J ( )d (1.16)

+1 1

where M is the total number of elements.


27

The integral functions (containing the kernels) can be lumped together in the functions Ai,j and Bi,j as follows: c(r)i (r) +M 1 j =0

j (r )Ai,j (r, r ) =

M 1 j =0

j (r ) Bi,j (r, r ) n

(1.17)

To form a set of linear algebraic equations, we take each node in turn as a load point r and perform the integrations indicated in Eq.(1.16). This will result in the following matrices: [ ] [A][] = [B ] (1.18) n where the matrices [A] and [B ] contain the integrals of the kernels normal rr |) derivative G(|n and the kernels G(|r r |) respectively, i.e. the functions Ai,j and Bi,j of Eq.(1.17). For 2D system, the fundamental solution for Laplace equation is equal: G(|r r |) = 1 1 1 1 ln = ln 2 |r r | 2 (x x )2 + (y y )2 (1.19)

The entries of the matrices are: Ai,j (r, r ) = +1 1 +1 1

G(|r r |) J ( )d n (1.20) G(|r r |)J ( )d

Bi,j (r, r ) =

where r depends on index i and r depends on index j . If R denotes the distance between point r and point r then: R = |r r | = (x x )2 + (y y )2 (1.21)

The normal derivative of Greens function in case of Laplace equation is: [ ] G(|r r |) G R G R x R y = = + = n R n R x n y n 1 [(x x)nx + (y y )ny ] (1.22) = 2R2

28 where:


R x x ; = x R x = nx ; n

R y y = y R (1.23) y = ny n

where nx and ny are dened by Eq.(1.10).

Integration of nonsingular integrals In case of nonsingular integrals, when point r and point r do not belong to the same boundary element, the standard GaussLegendre quadrature can be easily applicable to integrals of the general form: +1

f ( )d =1

g 1 i=0

wi f (i )

(1.24)

where g is the total number of Gaussian integration points, and i is the Gaussian coordinate with an associated weight function wi . The most frequently used values are listed in the Table 1.1 [18].

Integration of singular integrals The integrals can be calculated using Gauss quadrature rules for all elements except the one corresponding to the node under consideration. For this particular case the Ai,i coecients are equal zero and the Bi,i integrals can be calculated analytically. The distance R (see Eq.(1.21)) between point r and point r ( ) depends on local coordinate system in a following way: R( ) = |r r ( )| then Bi,i (R( )) =+1 1

(1.25)

L G(R( ))J ( )d = 2

+1

1

1 1 ln d 2 R( )

(1.26)

1.2. LAPLACE EQUATION Table 1.1: Onedimensional Gaussian quadraturen 4 6 i 1 2 1 2 3 1 2 3 4 5 1 2 3 4 5 6 1 2 3 4 5 6 7 8 Abscissa, i 0.33998104358485626480 0.86113631159405257522 0.23861918608319690863 0.66120938646626451366 0.93246951420315202781 0.14887433898163121089 0.43339539412924719080 0.67940956829902440623 0.86506336668898451073 0.97390652851717172008 0.12523340851146891547 0.36783149899818019375 0.58731795428661744730 0.76990267419430468704 0.90411725637047485668 0.98156063424671925069 0.09501250983763744019 0.28160355077925891323 0.45801677765722738634 0.61787624440264374845 0.75540440835500303390 0.86563120238783174388 0.94457502307323257608 0.98940093499164993260 Weight, wi 0.65214515486254614263 0.34785484513745385737 0.46791393457269104739 0.36076157304813860757 0.17132449237917034504 0.29552422471475287017 0.26926671930999635509 0.21908636251598204400 0.14945134915058059315 0.06667134430868813759 0.24914704581340278500 0.23349253653835480876 0.20316742672306592175 0.16007832854334622633 0.10693932599531843096 0.04717533638651182719 0.18945061045506849629 0.18260341504492358887 0.16915651939500253819 0.14959598881657673208 0.12462897125553387205 0.09515851168249278481 0.06225352393864789286 0.02715245941175409485

29

10

12

16

where in case of constant element J ( ) is dened by Eq.(1.14) and L denotes the length of the boundary element. This integral could easily be calculated analytically [23]. Hence Eq.(1.26) becomes, 1L Bi,i (R( )) = 2 ( ln 1L 2

) +1 (1.27)

For more complex cases a special logarithmically weighted numerical inte-

30

CHAPTER 1. TWODIMENSIONAL POTENTIAL PROBLEMSq4 i=4 q3 4 j=4 i=5 y j=5 r i=0 j=0 x 0 i=1 r() j=3 domaing=3

3

i=3 numerical integration points

rr

g=2 g=1 g=0

j=2

j=1

i=2

Figure 1.5: Numerical integration for constant boundary elements

gration formula can be used [21, 43, 45, 50, 54]: 0 1

f ( ) ln

1 d = wi f (i ) i=0

gl1

(1.28)

where: gl is the total number of logarithmic Gaussian integration points used and i is the Gaussian coordinate with an associated weight function wi (see Table 1.2). Note that the limits of integration are from 0 to 1 instead of the -1 to +1 range used in the nonsingular integrals of BIE.

1.2.3

Division of the boundary into linear boundary elements

Although, the constant element is a primitive one, in fact is a very convenient one. Particularly there is no problem with the corners of the region where the special treatment must be introduced to calculate the normal derivative of the state function, when the higher order interpolation is applied. Unlike

1.2. LAPLACE EQUATION Table 1.2: Onedimensional logarithmic Gaussian quadraturen 4 i 1 2 3 4 1 2 3 4 5 6 7 8 i 0.04144848019938322080 0.24527491432060225194 0.55616545356027583718 0.84898239453298517465 0.01332024416089246501 0.07975042901389493841 0.19787102932618805379 0.35415399435190941967 0.52945857523491727771 0.70181452993909996384 0.84937932044110667605 0.95332645005635978877 wi 0.38346406814513512485 0.38687531777476262734 0.19043512695014241536 0.03922548712995983245 0.16441660472800288683 0.23752561002330602050 0.22684198443191912637 0.17575407900607024499 0.11292403024675905186 0.05787221071778207239 0.02097907374213297804 0.00368640710402761901

31

8

the second order isoparametric quadratic element, the linear element has no signicant advantages over the constant element. Geometry of the region is interpolated in the same way. Only better interpolation of the unknowns is oered by this element, but the user has to pay for that with the corner troubles. Consider the region divided by the linear elements as it is shown in Fig. 1.6. At each node of the particular boundary element indicated by upper index (j ) two unknowns and q (j ) = are present. The same transformation n (as for constant element) is used to transfer the nodes from global to local coordinate system (see Fig. 1.7).

x( ) =

1 k=0

Nk ( )xk = N0 ( )x0 + N1 ( )x1 (1.29)

y ( ) =

1 k=0

Nk ( )yk = N0 ( )y0 + N1 ( )y1

32

CHAPTER 1. TWODIMENSIONAL POTENTIAL PROBLEMSq4 1111111111111111111111111111111111 0000000000000000000000000000000000 000000000000000000000 111111111111111111111 0000000000000000000000000000000000 1111111111111111111111111111111111 000000000000000000000 111111111111111111111 i=4 000000000000000000000 111111111111111111111 0000000000000000000000000000000000 1111111111111111111111111111111111 000000000000000000000 111111111111111111111 000000000000000000000 111111111111111111111 0000000000000000000000000000000000 1111111111111111111111111111111111 000000000000000000000 111111111111111111111 (j=4) 000000000000000000000 111111111111111111111 0000000000000000000000000000000000 1111111111111111111111111111111111 (j=3) 000000000000000000000 q111111111111111111111 000000000000000000000 111111111111111111111 q3 0000000000000000000000000000000000 1111111111111111111111111111111111 5 000000000000000000000 111111111111111111111 (j=4) (j=3) 000000000000000000000 111111111111111111111 0000000000000000000000000000000000 1111111111111111111111111111111111 000000000000000000000 111111111111111111111 4= 1111111111111111111111111111111111 000000000000000000000 111111111111111111111 0000000000000000000000000000000000 k=0= k=1 000000000000000000000 111111111111111111111 000000000000000000000 111111111111111111111 0000000000000000000000000000000000 1111111111111111111111111111111111 000000000000000000000 111111111111111111111 j=3 000000000000000000000 111111111111111111111 0000000000000000000000000000000000 1111111111111111111111111111111111 000000000000000000000 111111111111111111111 000000000000000000000 111111111111111111111 0000000000000000000000000000000000 (j=3) 1111111111111111111111111111111111 j=4 000000000000000000000 111111111111111111111 000000000000000000000 111111111111111111111 0000000000000000000000000000000000 1111111111111111111111111111111111 000000000000000000000 111111111111111111111 = 000000000000000000000 111111111111111111111 q4 5= k=1111111111111111111111 000000000000000000000 000000000000000000000 111111111111111111111 000000000000000000000 111111111111111111111 (j=4) 000000000000000000000 111111111111111111111 000000000000000000000 111111111111111111111

(j=4)

(j=3)

i=3

3

k=0

i=5

j=5

domain

j=2

i=0

j=0

i=2 j=1

i=1

Figure 1.6: Boundary discretization for a linear elements

N() N0()

N1() 1(j)

00000000000000000000000000 111111111111111111111111111 00000000000000000000000000 11111111111111111111111111 =+1 11111111111111111111111111 00000000000000000000000000 00000000000000000000000000 11111111111111111111111111 00000000000000000000000000 11111111111111111111111111 00000000000000000000000000 11111111111111111111111111 j 00000000000000000000000000 11111111111111111111111111 =0 00000000000000000000000000 11111111111111111111111111 (j) 00000000000000000000000000 11111111111111111111111111 0 00000000000000000000000000 11111111111111111111111111 00000000000000000000000000 11111111111111111111111111 00000000000000000000000000 11111111111111111111111111 00000000000000000000000000 11111111111111111111111111k=0 =1

k=1

Figure 1.7: Local coordinates of linear element number j

1.2. LAPLACE EQUATION For the linear element the shape functions are: 1 N( ) = [Nk=0 ( ), Nk=1 ( )]T = (1 + k ) 2 where: k=0 k = 1

33

(1.30)

(1.31) k=1 k = +1

The shape function dened by Eq.(1.30) are linear functions such as: Nk ( ) = 1 at the node k = 0 and Nk ( ) = 0 at the node k = 1, as it is shown in Fig. (1.7). Using the isoparametric linear elements the same basis interpolation functions are used for the solution variables representation: ( ) = and: ( ) k 0 1 Nk ( ) = = N0 ( ) + N1 ( ) n n n n k=01 1 k=0

Nk ( )k = N0 ( )0 + N1 ( )1

(1.32)

(1.33)

1.2.4


In case of linear boundary elements the numerical integration of the kernels can be done in a similar way as for constant elements. The Jacobian of transformation and the components of unit outward normal are calculated according to Eq.(1.14) and Eq.(1.10) respectively. The boundary curve is divided into boundary elements (segments) j . Now the BIE of Eq.(1.2) will take the form: M 1 G(|r r |) (j ) (r )d(r ) = c(r)i (r) + n j =0 j M 1 (j ) (r ) = G(|r r |) d(r ) (1.34) n j =0 j

34


The numerical integration is performed over each boundary element j using the local intrinsic coordinate , as follows: c(r)i (r) +M 1 +1 j =0 1 +1 1

=

M 1 j =0

G(|r r |) (j ) k (r )Nk ( )J ( )d = n k=01

G(|r r |)

1 (j ) (r ) k k=0

n

Nk ( )J ( )d

(1.35)

where M is the total number of quadratic elements. The nodal values are constant (do not depend on local coordinate ), so we can nally write Eq.(1.35) in the following form: c(r)i (r) +M 1 1 j =0 k=0 (j ) k (r ) (j )

+1 1

G(|r r |) Nk ( )J ( )d = n G(|r r |)Nk ( )J ( )d (1.36)

=

M 1 1 j =0 k=0

k (r ) n

+1 1

If we denote the terms containing the integrals of the kernels normal derirr |) vative G(|n and the kernels G(|r r |) as a and b respectively, we will get: +1 G(|r r |) (j ) ai,k (r, r ) = Nk ( )J ( )d (1.37) n 1(j ) bi,k (r, r )

=

+1 1

G(|r r |)Nk ( )J ( )d(j ) (j )

(1.38)

For the smooth boundary the integral functions ai,k and bi,k can be lumped together in the global functions Ai,j and Bi,j (see Eq.(1.40)) as follows: c(r)i (r) +M 1 1 j =0 k=0

ai,k (r, r )(r )k =(j ) (j )

M 1 1 j =0 k=0

bi,j (r, r )(j )

(r )k n

(j )

(1.39)

where r depends on index i and r depends on index j . To form a set of linear algebraic equations, we take each node in turn as a load point r and perform the integrations indicated in Eq.(1.36). This will

1.2. LAPLACE EQUATION result in the following matrices: [A][] = [B ] n [ ]

35

(1.40)

where the matrices [A] and [B ] are of the same size (only for the smooth boundary). The normal derivatives of Greens functions are calculated in the same way as for constant element (Eq.(1.22)). In case of the concave or convex corners a special treatment is needed.

The points r and r located in dierent elements Dealing rst with the second kernel G(|r r |) we have to consider two cases. The rst case is when the load and eld points are in dierent elements. Then integrals are not singular. And a bit more dicult problem when those two points are in the same element. In this case the singularity may occur. Let denote distance between two points as: R = |r r | = (x x)2 + (y y )2 (1.41)

then to calculate the integrals we have to consider the following two cases: The point r is the rst node (k = 0) of the linear element: R2 = [x ( ) x0 ] + [y ( ) y0 ] =2 2

= [N0 ( )x0 + N1 ( )x1 x0 ]2 + [N0 ( )y0 + N1 ( )y1 y0 ]2 (1.42) where: N0 ( ) and N1 ( ) are expressed by Eq.(1.52). So: [ R2

= [ +

1 1 (1 )x0 + (1 + )x1 x0 2 2 1 1 (1 )y0 + (1 + )y1 y0 2 2

]2 + ]2 (1.43)

36


The point r is a second node (k = 1) of the linear element: ]2 [ 1 1 2 R = (1 )x0 + (1 + )x1 x1 + 2 2 [ ]2 1 1 + (1 )y0 + (1 + )y1 y1 (1.44) 2 2 The points r and r are in the same element but r = r In this case, kernels are singular but the shape function Nk ( ) in the vicinity of r is of the order r. Therefore, the product of the kernels and the shape function is not singular, and the integrals can be evaluated using the standard Gaussian quadrature. So far, all the odiagonal coecients of the matrices [A] and [B ] have been calculated.

The points r and r are in the same linear element but r = r so R0 In this case, the standard Gaussian quadrature cannot be used because of 1 the singularity of the kernels. Dealing with the kernel G(r, r ) = 21 ln R 1 rst, it is clear that as r coincides with r , the singularity is of the form ln as 0. Fortunately, this form of integral can be calculated by using the special logarithmic Gaussian quadrature scheme [21,43,45,50,54,85,86] given bellow: 1 gl1 1 f ( ) ln d = wi f (i ) (1.45) 0 i=0 where: gl is the total number of logarithmic Gaussian integration points used and i is the Gaussian coordinate with an associated weight function wi (see Table 1.2). Note that the limits of integration are now from 0 to 1 instead of the -1 to +1 range used in the nonsingular integrals. A simple linear transformation can be used to transform the integral variable from to as follows:

1.2. LAPLACE EQUATION a) k=1 (0)

37 b) k=1 =+1 (1) =0

=+1 (0) =1

=0 k=0 =1 (0) =0 k=0 (1) =1 (1) =1

=0

Figure 1.8: Coordinates transformation. a) r is in the rst node of the element, b) in the second node

1. if the position vector r is the rst node (k=0) of the element: (k=0) = 0.5(1 + ) 2. if the position vector r is the second node (k=1) of the element: (k=1) = 0.5(1 ) (1.47) (1.46)

For the rst node the Eq.(1.43) can be rearranged in the following way: [ ] ( )2 R2 = [0.5(1 + )]2 (x1 x0 )2 + (y1 y0 )2 = (0) L2 where L is a length of linear boundary element. For the second node the Eq.(1.44) can be rearranged in the following way: [ ] ( )2 R2 = [0.5(1 )]2 (x0 x1 )2 + (y0 y1 )2 = (1) L2 (1.49) (1.48)

Therefore, a general expression can be written for the logarithmic term as follows: G(|r r |) = 1 1 1 1 1 1 1 ln = ln (k) = ln (k) ln L (1.50) 2 |r r | 2 L 2 2

38


where: k indicates the node number and (k) transforms the integration limits from (1 to +1) to (0 to +1) for k = 0 and from (+1 to 1) to (0 to +1) for k = 1. Examining closely the Green function (Eq.(1.50)) as r approaches r , this function can be divided into two distinct parts: logarithmic part and a nonlogarithmic one (see Eq.(1.50)). Dealing now with the rst kernel (normal derivative of the Green function), it can be shown that it contains terms of order (1k) as (k) 0. Therefore, we can no longer use the Gaussian quadrature technique, even if a very large number of Gaussian points are used. Furthermore, we also need to explicitly calculate the parameter c(r) because its contribution is added to the diagonal terms of the [A] matrix. However, because all nondiagonal coecients of the [A] matrix can be calculated, there is a way to overcome this problem, which will be discussed later.

1.2.5

Division of the boundary into quadratic boundary elements

Particularly in optical or in impedance tomography we have to deal with very complicated shapes of the regions. That is why the main attention will be devoted to the second order interpolation over boundary elements in 2D and 3D space. Let consider the region presented in Fig. 1.9. We can dene a new coordinate system that is local to the element using an intrinsic variable , with its origin at the midpoint of the element and values of -1 and +1 at the end nodes as it is shown in Fig. 1.10. x( ) =2 k=0

Nk ( )xk = N0 ( )x0 + N1 ( )x1 + N2 ( )x2 (1.51)

y ( ) =

2 k=0

Nk ( )yk = N0 ( )y0 + N1 ( )y1 + N2 ( )y2

where Nk ( ) are quadratic functions such as: Nk ( ) = 1 at its own node for

1.2. LAPLACE EQUATIONk=2 0000000000000000 1111111111111111 1111111111111111111111111111111 0000000000000000000000000000000 0000000000000000 1111111111111111 0000000000000000000000000000000 1111111111111111111111111111111 (j=3) 0000000000000000 1111111111111111 0000000000000000000000000000000 q 7 = q k=1 0000000000000000 1111111111111111 q 1111111111111111111111111111111 = q(j=4) 0000000000000000000000000000000 1111111111111111111111111111111 i=8 0000000000000000 1111111111111111 10 k=2 0000000000000000000000000000000 1111111111111111111111111111111 0000000000000000 1111111111111111 0000000000000000000000000000000 1111111111111111111111111111111 0000000000000000 1111111111111111 i=9 0000000000000000000000000000000 1111111111111111111111111111111 i=7 0000000000000000 1111111111111111 (j=3) (j=4) 0000000000000000000000000000000 1111111111111111111111111111111 = 0000000000000000 1111111111111111 = k=2 j=3 k=0 0000000000000000000000000000000 1111111111111111111111111111111 8 (j=3) (j=2) 0000000000000000 1111111111111111 0000000000000000000000000000000 1111111111111111111111111111111 q = q k=0 = q k=2 0000000000000000 1111111111111111 j=4 0000000000000000000000000000000 1111111111111111111111111111111 6 0000000000000000 1111111111111111 0000000000000000000000000000000 1111111111111111111111111111111 0000000000000000 1111111111111111 i=6 0000000000000000000000000000000 1111111111111111111111111111111 (j=3) (j=2) 0000000000000000000000000000000 1111111111111111111111111111111 6= k=0 = k=2 0000000000000000000000000000000 1111111111111111111111111111111 i=10 (j=4) 8

39q = q k=0= q(j=3)(j=4)

q9= q(j=4) k=1

10= k=2

i=11

j=5

domain

j=2

i=5 i=4

j=0 i=0 i=1 i=2

j=1 i=3

Figure 1.9: Boundary discretization for a quadratic boundary elements k=2 =+1 endpoint k=1 =0 midpoint

k=0 =1 endpoint

Figure 1.10: An isoparametric quadratic element

example the node k = 0 and Nk ( ) = 0 at the other two nodes k = 1 and k = 2, resulting in the following: N0 ( ) = (1 ) = 0.5 ( 1) 2 N1 ( ) = (1 + )(1 ) = 1 2 N2 ( ) = + (1 + ) = 0.5 ( + 1) 2

(1.52)

40

CHAPTER 1. TWODIMENSIONAL POTENTIAL PROBLEMS1

N0

0.5

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

N1

0.5

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

N2

0.5

0 1 0.8 0.6 0.4 0.2

0

0.2

0.4

0.6

0.8

1

Figure 1.11: Basis interpolation functions distribution

Using the isoparametric elements the same basis functions are used for the solution variables:2 k=0

( ) =

Nk ( )k = N0 ( )0 + N1 ( )1 + N2 ( )2

(1.53)

and: k 0 1 2 ( ) = Nk ( ) = N0 ( ) + N1 ( ) + N2 ( ) n n n n n k=02

(1.54)

The basis functions distribution over the boundary element are presented in Fig. 1.11.

1.2. LAPLACE EQUATIONq k=0= q q9= q(j+1) k=2 111111111111111111111111111111 000000000000000000000000000000 0000000000000000 1111111111111111 k=1 000000000000000000000000000000 111111111111111111111111111111 0000000000000000 1111111111111111 000000000000000000000000000000 111111111111111111111111111111 0000000000000000 1111111111111111 000000000000000000000000000000 111111111111111111111111111111 0000000000000000 1111111111111111 q(j) q = q(j+1) 000000000000000000000000000000 k=1 0000000000000000 1111111111111111 10 111111111111111111111111111111 k=2 000000000000000000000000000000 111111111111111111111111111111 0000000000000000 1111111111111111 000000000000000000000000000000 111111111111111111111111111111 k=2 0000000000000000 01111111111111111 000000000000000000000000000000 111111111111111111111111111111 0000000000000000 1111111111111111 (j+1) k=1 000000000000000000000000000000 111111111111111111111111111111 0000000000000000 = (j) k=0 1111111111111111 000000000000000000000000000000 111111111111111111111111111111 k=2 0000000000000000 1111111111111111 1 000000000000000000000000000000 111111111111111111111111111111 0000000000000000 1111111111111111 j q(j) 000000000000000000000000000000 111111111111111111111111111111 k=0 0000000000000000 1111111111111111 000000000000000000000000000000 111111111111111111111111111111 j+1 k=0 0000000000000000 1111111111111111 000000000000000000000000000000 111111111111111111111111111111 (j1) (j) 000000000000000000000000000000 111111111111111111111111111111 k=2 = k=0 000000000000000000000000000000 111111111111111111111111111111 (j+1) (j+2) 2 q k=2 9 k=1 0000000000000 1111111111111 111111111111111111111111111111 000000000000000000000000000000 0000000000000 1111111111111 000000000000000000000000000000 111111111111111111111111111111 0000000000000 1111111111111 000000000000000000000000000000 111111111111111111111111111111 0000000000000 1111111111111 000000000000000000000000000000 111111111111111111111111111111 q(j) 0000000000000 1111111111111 000000000000000000000000000000 111111111111111111111111111111 q = q(j+1) k=1 10 k=2 0000000000000 1111111111111 000000000000000000000000000000 111111111111111111111111111111 0 0000000000000 1111111111111 000000000000000000000000000000 111111111111111111111111111111 k=2 0000000000000 1111111111111 000000000000000000000000000000 111111111111111111111111111111 (j+1) 0000000000000 000000000000000000000000000000 111111111111111111111111111111 k=1 k=0 =1111111111111 (j) 1 0000000000000 1111111111111 000000000000000000000000000000 111111111111111111111111111111 k=2 q(j) k=0 0000000000000 1111111111111 000000000000000000000000000000 111111111111111111111111111111 j 0000000000000 1111111111111 000000000000000000000000000000 111111111111111111111111111111 j+1 k=0 0000000000000 1111111111111 000000000000000000000000000000 111111111111111111111111111111 000000000000000000000000000000 111111111111111111111111111111 (j1) 000000000000000000000000000000 111111111111111111111111111111 2 k=2 = (j) k=0 k=2 = (j+2) k=0(j+1) (j+1) (j)

41

k=2 = k=0

q =q(j+1)

q(j+1) k=0

(j)

Figure 1.12: Variation of q = along the boundary line interpolated by qun adratic boundary elements without (upper) and with (lower) jumps between elements

1.2.6


In case of quadratic boundary elements the numerical integration of the kernels could be done in a similar way as for constant and linear elements. The Jacobian of transformation and the components of unit outward normal are calculated according to Eq.(1.14) and Eq.(1.10) respectively. The components of unit outward normal (see Eq.(1.10)) are functions of local coordinate : [ ] 1 dy ( ) nx ( ) = J ( ) d [ ] 1 dx( ) ny ( ) = J ( ) d

(1.55)

42 where:


dN0 ( ) dN1 ( ) dN2 ( ) dx( ) = x0 + x1 + x2 d d d d (1.56) dy ( ) dN0 ( ) dN1 ( ) dN2 ( ) = y0 + y1 + y2 d d d d and the dierentials of the shape functions are easily determined as follows: ( ) dN0 ( ) d 1 (1 ) = = d d 2 2 dN1 ( ) d = ((1 + )(1 )) = 2 (1.57) d d ( ) dN2 ( ) d 1 = + (1 + ) = + d d 2 2 The numerical integration is performed over each boundary element j using the local intrinsic coordinate , as follows: 2 M 1 +1 G(|r r |) (j ) k (r )Nk ( ) c(r)i (r) + J ( )d = n j =0 1 k=0 2 M 1 +1 (j ) k (r ) = G(|r r |)Nk ( )J ( )d (1.58) n 1 j =0 k=0 where M is the total number of quadratic elements. The nodal values are constant, so we can nally write Eq.(1.58) in the following form: +1 M 1 2 G(|r r |) (j ) c(r)i (r) + k (r ) Nk ( )J ( )d = n 1 j =0 k=0 M 1 2 (j ) k (r ) +1 G(|r r |)Nk ( )J ( )d (1.59) = n 1 j =0 k=0 If we denote the terms containing the integrals of the kernels normal derirr |) and the kernels G(|r r |) as a and b respectively, we will vative G(|n get: +1 G(|r r |) (j ) Nk ( )J ( )d (1.60) ai,k (r, r ) = n 1

1.2. LAPLACE EQUATION +1 (j ) bi,k (r, r ) = G(|r r |)Nk ( )J ( )d1 (j ) (j )

43 (1.61)

For the smooth boundary the integral functions ai,k and bi,k can be lumped together in the global functions Ai,j and Bi,j (see Eq.(1.63)) as follows: c(r)i (r) +M 1 2 j =0 k=0 (j ) (j ) ai,k (r, r )(r )k

=

M 1 2 j =0 k=0

(j ) (j ) (r )k bi,j (r, r )

n

(1.62)

where r depends on index i and r depends on index j . To form a set of linear algebraic equations, we take each node in turn as a load point r and perform the integrations indicated in Eq.(1.60) and in Eq.(1.61). This will result in the following matrices: [ ] [A][] = [B ] (1.63) n where the matrices [A] and [B ] are of the same size (only for the smooth boundary). The normal derivatives of Green functions are calculated in the same way as for constant and linear elements ( see Eq.(1.22)). In case of the concave or convex corners, a special treatment is needed.

The points r and r located in dierent elements Dealing rst with the second kernel G(|r r |) we have to consider two cases. The rst case when load and eld points are in dierent elements. Then integrals are not singular. And a bit more dicult problem when those two points are in the same element. In this case the singularity may occur. If distance between two points we will denote as: R = |r r | = (x x)2 + (y y )2 (1.64)

then to calculate the integrals we have to consider the following three cases:

44


The point r is the rst node (k = 0) of the element: R2 = [x ( ) x0 ] + [y ( ) y0 ] =2 2

= [N0 ( )x0 + N1 ( )x1 + N2 ( )x2 x0 ]2 + + [N0 ( )y0 + N1 ( )y1 + N2 ( )y2 y0 ]2

(1.65)

where: N0 ( ), N1 ( ) and N2 ( ) are expressed by Eq.(1.52). So: R2

[ ]2 = (1 )x0 + (1 + )(1 )x1 + (1 + )x2 x0 + 2 2 [ ]2 (1.66) + (1 )y0 + (1 + )(1 )y1 + (1 + )y2 y0 2 2

The point r is second node (k = 1) of the element: R2

[ ]2 = (1 )x0 + (1 + )(1 )x1 + (1 + )x2 x1 + 2 2 [ ]2 + (1 )y0 + (1 + )(1 )y1 + (1 + )y2 y1 (1.67) 2 2

The point r is third node (k = 2) of the element: R2

[ ]2 = (1 )x0 + (1 + )(1 )x1 + (1 + )x2 x2 + 2 2 [ ]2 + (1 )y0 + (1 + )(1 )y1 + (1 + )y2 y2 (1.68) 2 2

The points r and r are in the same element but r = r In this case, kernels are singular but the shape function Nk ( ) in the vicinity of r is of the order r. Therefore, the product of the kernels and the shape function is not singular, and the integrals can be evaluated using the standard Gaussian quadrature. So far, all the odiagonal coecients of the matrices [A] and [B ] have been calculated.

1.2. LAPLACE EQUATION The points r and r are in the same element but r = r , so R 0

45

In this case, the standard Gaussian quadrature cannot be used because of 1 the singularity of the kernels. Dealing with the kernel G(r, r ) = 21 ln R rst, it is clear that as r coincides with r , the singularity is of the form 1 as 0. Fortunately, this form of integral can be calculated by using ln the special logarithmic Gaussian quadrature scheme [21, 43, 45, 50, 54] given bellow: 1 gl1 1 wi f (i ) (1.69) f ( ) ln d = 0 i=0 where: gl is the total number of logarithmic Gaussian integration points used and i is the Gaussian coordinate with an associated weight function wi (see Table 1.2). Note that the limits of integration are now from 0 to 1 instead of the 1 to +1 range used in the nonsingular integrals.

k=2 =+1 =1 =0 =1 k=0 =0(0) (0)

(0)

=+1(1)

(1) =+1

=1 =0 =0

=0

(2)

k=1

=1 (1)

=0 = 0 =1(1)

(1)

= 1

(1)

(2)

=1

(2)

Figure 1.13: Coordinates transformation. a) r is in the rst node of the element, b) second node and c) third node

A simple linear transformation can be used to transform the integral variable from to as follows: 1. if the position vector r is the rst node (k = 0) of the element: (k=0) = 0.5(1 + ) (1.70)

46

CHAPTER 1. TWODIMENSIONAL POTENTIAL PROBLEMS 2. if the position vector r is the second node (k = 1) of the element - the element is divided into two subelements: (k=1) = for 1 < < 0 and (k=1) = for 0 < < 1 (1.71) 3. if the position vector r is the third node (k = 2) of the element: (k=2) = 0.5(1 ) (1.72)

For the rst node the Eq.(1.66) can be rearranged in the following way: { R2 = [0.5(1 + )]2 [( 2)x0 + 2(1 )x1 + x2 ]2 + (1.73) [ } ( ) ( ) ( )2 ] 2 2 (0) (0) + [( 2)y0 + 2(1 )y1 + y2 ]2 = (0) fx ( ) + fy ( ) For the second node the Eq.(1.67) can be rearranged in the following way: { R2 = 2 [0.5( 1)x0 x1 + 0.5( + 1)x2 ]2 + (1.74) [( } ( ) ) ( )2 ] 2 2 (1) (1) + [0.5( 1)y0 y1 + 0.5( + 1)y2 ]2 = (1) fx ( ) + fy ( ) For the third node the Eq.(1.69) can be rearranged in the following way: { R2 = [0.5(1 )]2 [x0 + 2( + 1)x1 2( + 2)x2 ]2 + (1.75) [ } ( ) ( ) ( )2 ] 2 2 (2) 2 (2) (2) + [y0 + 2( + 1)y1 2( + 2)y2 ] = fx ( ) + fy ( ) Therefore, a general expression can be written for the logarithmic term as follows: G(|r r |) = 1 1 1 ln = ln 2 |r r | 2 1 )2 ( )2 = ( k ) ( k ) (k) fx ( ) + fy ( ) 1 1 1 [( (k) )2 ( (k) )2 ] 1 (1.76) ln (k) ln fx ( ) + fy ( ) = 2 2 2 (

where: k indicate, the node number and (k) transforms the integration limits from (1 to +1) to (0 to +1) and changes its value according to the position of r on the element. As for linear boundary element the second kernel as r approaches r , can be divided into two distinct parts: logarithmic part and a nonlogarithmic one (see Eq.(1.76)). Dealing now with the rst kernel, it can


47

1 be shown that it contains terms of order as 0. Therefore, we can no longer use the Gaussian quadrature technique, even if a very large number of Gaussian points are used. Furthermore, we also need to explicitly calculate the parameter c(r) because its contribution is added to the diagonal terms of the [A] matrix. However, because all nondiagonal coecients of the [A] matrix can be calculated, there is a way to overcome this problem. Next section is devoted to this particular problem.

1.2.7

Numerical integration of the c(r) coecient

We have two sets of points: points r where the unit sources are applied and points r where we have to satisfy boundary conditions. The problem is that some integrals in Eq.(1.59) only exist in the sense of a limiting value as r approaches r . This is explained in Fig. 1.14 for twodimensional potential problems [15]. We dene a region of exclusion around point r, with radius and we integrate around it. The integrals in Eq.(1.59) can now be split up into integrals over S S that are, the part of curve without the exclusion zone and into integrals over s that is, the circular boundary. As goes to zero it does not matter if we integrate over s or S as shown in Fig. 1.14. The lefthand side coecient c(r) of the Eq.(1.59) for 2D problems could be written as: c(r) = 1 2 (1.77)

where (see Fig. 1.15) is dened as the angle subtended at r by s in the limit as tends to zero. For 3D problems the same procedure could be applied (as it is sketched in Fig. 1.16) but for general corners expression similar to Eq.(1.77) is more complicated [15]. The function c(r) does not have to be calculated explicitly, and can be obtained indirectly by utilizing some physical considerations. It is based on the fact that the Boundary Integral Equations (BIE) matrices must apply to any physical problem with a unique solution. Any physical problem can be chosen as long as the solution does not depend on the geometry. The simplest issue to consider is the case where the potential is constant throughout the solution domain because that

48


d r y r

n

s

SS

S

0

x

Figure 1.14: For twodimensional potential problem semicircles around the boundary point r

leads to zero potential gradients everywhere. Therefore the righthand side of Eq.(1.59) becomes zero. Since the values are the same, the sum of all coecients in any row of the lefthand side matrices must be zero. The diagonal term can be determined as the sum of the nondiagonal ones as follows: Ai,j = N j =0

Ai,j

for

i = 0, 1, 2, . . . , N

(1.78)

j =i

where the i and j are the row and column counters respectively, while the N is the total nodes number.

1.2.8

Internal points calculation

Very close to the problem of singularities mentioned in the previous section is the problem of the so called quasisingularity, which occurs in the Boundary Element Method (BEM) analysis. Namely, when calculating the internal points placed near the boundary, one faces the small arguments of the kernels.


49

r y r n

s

0

x

SS

Figure 1.15: Boundary point r is located in the corner

000 111 d d 0000000000000000000000000000000000000000 1111111111111111111111111111111111111111 0000000000000000000000000000000000000000 1111111111111111111111111111111111111111 ns

000 111

1111111111111111111111111111111111111111 0000000000000000000000000000000000000000 0000000000000000000000000000000000000000 1111111111111111111111111111111111111111 0000000000000000000000000000000000000000 1111111111111111111111111111111111111111 0000000000000000000000000000000000000000 1111111111111111111111111111111111111111 0000000000000000000000000000000000000000 1111111111111111111111111111111111111111 0000000000000000000000000000000000000000 1111111111111111111111111111111111111111 SS 0000000000000000000000000000000000000000 1111111111111111111111111111111111111111 0000000000000000000000000000000000000000 1111111111111111111111111111111111111111 0000000000000000000000000000000000000000 1111111111111111111111111111111111111111 0000000000000000000000000000000000000000 1111111111111111111111111111111111111111 0000000000000000000000000000000000000000 1111111111111111111111111111111111111111 0000000000000000000000000000000000000000 1111111111111111111111111111111111111111 0000000000000000000000000000000000000000 1111111111111111111111111111111111111111 0000000000000000000000000000000000000000 1111111111111111111111111111111111111111Figure 1.16: For the threedimensional potential problem hemisphere around the boundary point

Due to this reason the numerical integration may be not precise enough as we can observe in Fig. 1.17. To avoid this situation some of the authors [49] propose to use the classical technique consisting of adding and subtracting a part of the integrand, which produces a quasisingularity. The others [14] suggest that instead of

50


Figure 1.17: Internal points calculation for 16 (left) and 32 (right) boundary elements discretization division the kernel into singular and nonsingular parts, simply multiplying 1 and dividing it by ln as follows: 0 1

f ( )d =0

1

1 f ( ) ln 1 ln

d =0

1

1 wi h(i ) h( ) ln d = i=0

gl1

(1.79)

The righthand side of the above equation is exactly as required by the logarithmic Gaussian quadrature. This latter approach is much simpler to program and is only slightly less accurate than the former approach of dividing the kernel into singular and nonsingular parts. However, such a problem could be avoided when the number of elements increases as it is presented in Fig. 1.17. For the internal points the function c(r) is equal one, so the state values could be calculated as follows: G(|r r |) (r ) (r) = (r )d(r ) + G(|r r |) d(r ) (1.80) n n It is worth to mention that for internal points usually more Gaussian integration points are demanded to achieve satisfactory precision. In this particular case, for the boundary variables the four points Gaussian quadrature integration rule was applied when for the internal point variables the ten points Gaussian quadrature integration rule was adopted. That makes integration a very timeconsuming process, so some authors recommend to avoid calculation variables in the internal points and plan division of the structure for some substructures in such a way that the interesting us internal points could be treated as the points belonging to the interface between substructures.


51

The gradient components at point r in the x and y directions are computed by taking the derivative of Eq.(1.80) in x and y direction respectively: (r) = x (r ) d G ( | r r | ) n x (1.81) ( ) (r ) (r) G(|r r |) = (r )d + G(|r r |) d y n n y y

x

(

G(|r r |) n

)

(r )d +

Where the derivatives of the kernel x ( G(|r r |) n )

G n

are well known and are given as: ( G(r, r ) n ) = 1 y y (1.82) 2 R 2

1 x x = and 2 2 R y

The derivatives of the kernel G are a little bit more complicated: [ ] G(|r r |) 1 x x y y = nx + ny = x 2 x R2 R2 [ ( )] nx 2(x x) x x 1 y y 2 + nx + ny (1.83) = 2 x R R2 R2 R2 [ ] 1 x x G(|r r |) y y = nx + ny = y 2 y R2 R2 [ ( )] 1 ny 2(y y ) x x y y = 2 + n x + ny (1.84) 2 y R R2 R2 R2

1.2.9

Symbolic calculation

Symbolic calculation of matrix entries became very popular in the Finite Element Method (FEM). It speeds up the calculation and improves precision as well. However, for the BEM it is much more dicult due to the fact that we have much more complicated integrals. But even though it is still possible for some of those integrals. Symbolic integration example for non singular integrals in case of Laplaces equation approximated by second order boundary elements is presented below.

52


a_aux[0]=(yp*(0.07957747154594767*cordx[0] - 0.1061032953945969*cordx[1] +0.026525823848649224*cordx[2]))/ (xp*xp - 2.*xp*xq + xq*xq + yp*yp - 2.*yp*yq + yq*yq) + (0.07957747154594767*yq*cordx[0] - 0.1061032953945969*yq*cordx[1] + 0.026525823848649224*yq*cordx[2] + 0.07957747154594767*xp*cordy[0] - 0.07957747154594767*xq*cordy[0] - 0.1061032953945969*xp*cordy[1] + 0.1061032953945969*xq*cordy[1] +0.026525823848649224*xp*cordy[2] - 0.026525823848649224*xq*cordy[2])/ (-xp*xp + 2.*xp*xq - xq*xq - yp*yp + 2.*yp*yq - yq*yq);

a_aux[1]=(0.1061032953945969*(yp*(cordx[0] - cordx[2]) + yq*(-cordx[0] + cordx[2]) - (xp - xq)*(cordy[0] - cordy[2])))/ (xp*xp - 2.*xp*xq + xq*xq + yp*yp - 2.*yp*yq + yq*yq);

a_aux[2]= (0.026525823848649224*yp*cordx[0] - 0.026525823848649224*yq*cordx[0] - 0.1061032953945969*yp*cordx[1] + 0.1061032953945969*yq*cordx[1] + 0.07957747154594767*yp*cordx[2] - 0.07957747154594767*yq*cordx[2] - 0.026525823848649224*xp*cordy[0] + 0.026525823848649224*xq*cordy[0] + 0.1061032953945969*xp*cordy[1] - 0.1061032953945969*xq*cordy[1] - 0.07957747154594767*xp*cordy[2] + 0.07957747154594767*xq*cordy[2])/ (-xp*xp + 2.*xp*xq - xq*xq - yp*yp + 2.*yp*yq - yq*yq);

where: xp, yp are equivalent to x, y and xq , yq are equivalent to x , y respectively, cordx and cordy denotes coordinates of the nodes as it is shown for example in Fig.1.10, and a_aux stands for the coecients ai,k - consult Eq.(1.60). As we can see the above expressions are so complicated that such approach cannot be recommended on the stage of software prototyping. So far it is dicult to judge how much the symbolic integration will help to speed up calculation and improve the precision.

1.3. DIFFUSION EQUATION

53

1.3

Diusion equation

For the homogeneous regions and a steady state the diusion equation could be written as follows: D2 (r) + (r) = 0 or: 2 (r) (r) = 0 D (1.86) (1.85)

where k 2 = D is so called wave number and for the steady state, in some cases (neutron transport problem) is a real number [42].

For optical tomography, usually diusion coecient D = 0.03 cm and = 0.1 cm1 , so k = D = 1.8257 cm1 . 2 (r) k 2 (r) = 0 (1.87)

The boundary integral equation corresponding to Eq.(1.87) can be written in the form [21] G(|r r |) (r ) c(r)(r) + D (r )d(r ) = D G(|r r |) d(r ) n n (1.88) For 2D system, a modied Bessel function of the second kind and zero order, can be chosen as a fundamental solution [21] G(|r r |) = where R = |r r | = 1 K0 (k R) 2D (1.89)

(x x )2 + (y y )2

rr |) To calculate the rst kernel G(|n , the function G(|r r |) is dierentiated with respect to the unit outward normal at the point r : ) ( G(|r r |) 1 = K0 (k R) = n n 2D (1.90) ) ( ) ( R k R 1 = K0 (k R) = K1 ( k R ) R 2D n 2D n

54


where: K1 is the modied Bessel function of the second kind and rst order [2]. The derivative of the distance R with respect to the unit outward normal n = 1x + 1y at the point r is as follows: R R x R y R R = + = nx + ny n x n y n x y where: R x x = x R and R y y = y R (1.92) (1.91)

Therefore, the kernel from Eq.(1.90) can be expressed: ( )( ) G(|r r |) k x x y y = K1 ( k R ) nx + ny n 2D R R

(1.93)

So the equation (1.88) becomes: )( ) ( x x y y k c(r)(r) + D K1 (k R) nx + ny (r )d = 2D R R (r ) 1 K0 (k R) d (1.94) = D n 2D The boundary of the solution domain is divided into a number of connected elements j . Over each element, the variation of the geometry and variables must be described. Let assume, that as in case of Laplace equation the isoparametric quadratic element will be used. The BIE can be solved numerically by dividing the boundary into elements using appropriate shape functions in local coordinate system, the same as in Laplace equation section. Than Eq.(1.94) will take the form: c(r)i (r) + M 1 2 k 2j =0 l=0 M 1 2 j =0 l=0

(j ) l (r ) (j )

+1

( K1 (k R)

1

) y y x x nx + ny Nl ( )J ( )d R R (1.95)

1 = 2

l (r ) n

+1

1

K0 (k R)Nl ( )J ( )d

where: M is the total number of elements. Now k denotes the wave number (see Eq.(1.87)) and l represents the local number of the nodes of particular boundary element (j ).

1.3. DIFFUSION EQUATION

55

1.3.1

Treatment of singularity

For the small arguments x n , the modied Bessel function becomes, asymptotically simple power (see Eq.(1.97)) of their arguments [2]: for n = 0: 1 K0 (x) = ln(x) = + ln x and for n > 0: (n 1)! ( x )n Kn (x) = 2 2 So for the rst order we will get:80BesselK(0,x) ln(x) 4 3.5 3 2.5

(1.96)

(1.97)

4.5

70 60 50 40

BesselK(1,x) 1./x

2

301.5 1 0.5 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

20 10 0 0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Figure 1.18: Comparison between the modied Bessel function of the second kind zero order (left) and the rst order (right) and the function 1/x for the arguments less than 1 (1 1)! ( x )1 1 K1 (x) = (1.98) = 2 2 x Taking above into account, the Green functions for the small arguments could be written as follows: 1 1 1 K0 (k R) ( ln(k R)) = (ln(k ) + ln(R)) = G(|r r |) = = 2D 2D 2D 1 1 = ln(k ) + ln 2D (l) (l) ( l ) 2 2 fx ( ) + fy ( ) ] [ ) 1 1 1 ( (l ) 2 2 (l ) = ln(k ) ln fx ( ) + fy ( ) + ln (l) 2D 2 =

(1.99)

56


where k is the wave number, superscript (l) denotes the number of the node in particular boundary element, is local coordinate system (see the Fig. 1.13) and fx and fy are the auxiliary functions for the rst time introduced in Eq.(1.73). The rst two terms of Eq.(1.99) are nonlogarithmic and the last one is the logarithmic type. That is why this kernel needs a special treatment (see the chapter 1.2.2 which is devoted to the Laplace equation). ( ) G(|r r |) k R k 1 x x x y y y = = K1 ( k R ) + = n 2D n 2D k R R n R n [ ] 1 x y = (x x ) + (y y ) (1.100) 2DR2 n n

1.3.2


After solving the system of linear equations, all the values of potential and potential gradient are determined on the boundary. So, it is straightforward matter to determine the potentials and gradient components in x and y direction, at any interior point. Because the interior point never lies on the boundary, there is no possibility of r and r coinciding and the kernels are no longer singular. )( ) ( k x x y y (r) = K1 (k R) nx + ny (r )d + 2 R R (r ) 1 K0 (k R) d (1.101) + n 2 (r) = x +

x

)( )] [( x x y y k nx + ny (r )d + K1 (k R) 2 R R (1.102)

1 K0 (k R) (r ) d 2 x n

In the similar way the y component of the gradient can be calculated. It is worth to mention that all dierentiations are made in the point r not in the point r .

1.4. DIFFUSION EQUATION IN FREQUENCY DOMAIN

57

1.3.3

Example

As an example let consider the steady state for the diusion equation (for example Eq.(1.85)) in a square region of dimension a for Dirichlet boundary conditions: (x = 0) = 0 and (x = a) = 10 and Neumann boundary = = 0. = conditions: n y yy =0 y =a

So in reality, this problem is 1D, but formally it can be solved in 2D space. Comparison of the results with the analytical solution for two dierent discre for dierent tizations is presented in Fig. 1.19. But the relative error of n zero order boundary elements density discretization is presented in Fig. 1.20. We can observe an exponential error decreasing with respect to the number of elements. Also it is very interesting to compare the results of two numerical method like FEM and BEM with the analytical solution. One can easily notice that for FEM the error achieves enormously big values, when for BEM the error remains constant within the region and is equal to about 2 % (see Fig. 1.21). This is an immanent and very positive feature of the Boundary Element Method.

1.4

Diusion equation in frequency domain

The time dependent diusion equation for potential problems can be written as follows: D2 (r, t) + (r, t) =0 t (1.103)

where r represents space dimensions and t represents time. Material properties assumed constant in space and time were described in previous section. Initial values of (r, t) must be prescribed at time t = t0 or zero, as follows: (r, t) |t=t0 = (r, t0 ) = const.

5810 10 10 10 10 10 10 10 101 0

CHAPTER 1. TWODIMENSIONAL POTENTIAL PROBLEMS101

Analytic BEM

10 10 10 10 10 10 10 10

0

Analytic BEM

1

1

2

2

3

3

4

4

5

5

6

6

7

7

0

1

2

3

4

5

6

7

8

9 x

10

0

1

2

3

4

5

6

7

8

9 x

10

Figure 1.19: Solution for 16 boundary elements (left) and for 128 boundary elements (right)50

40 [ %]

30

20

10

0

16 32

64

96

128 number of elements

256

Figure 1.20: For the zero order boundary elements the relative error for versus number of elements100

n

16 14

80

12 10

60 error [%]

error [%]

8 6 4 2 0 2

FEM BEM

40

20

0

20 0

2

4

x [mm]

6

8

10

4 0

2

4

x [mm]

6

8

10

Figure 1.21: Error comparison for FEM (right) and BEM (left) in case of the example presented in subsection 1.3.3


59

The boundary conditions are usually in the form of a prescribed state function on part of the boundary 1 and prescribed state (potential) gradient function on another part of the boundary 2 as follows: (r, t) = h(r, t) on 2 (1.104) t where the functions g (r, t) and h(r, t) are known as functions in time. (r, t) = g (r, t) on 1 ; The Laplace transform is used to remove the time dependency of the integral equations. To obtain the physical quantities, an inverse Laplace transform is applied. The Laplace transform can be dened as follows [14]: L [(r, t)] = (r, t)est dt = L (r, s) (1.105)0

where s is called the transform parameter. Therefore, using the properties of Laplace transforms, the dierential equation can be written as follows: 1 2 L (r, t) s L (r, t) = 0 (r, t0 ) (1.106) D D where k 2 = s is so called wave number. D Note that the boundary conditions in the form of the functions g and h (Eq. (1.104)) also have to be expressed as Laplace transforms. It can be veried that the fundamental solution for this dierential equation is given by the following expression (see [14]): ( ) 1 s GL (|r r | , s) = K0 R (1.107) 2D D where K0 is the modied Bessel function of the second kind of zero order. Using the same procedure as that outlined in case of Laplace equation for twodimensional potential problems, we can use Greens second identity to arrive at a boundary integral equation as follows: GL (|r r | , s) L (r , s)d = c(r)L (r, s) + D n (1.108) L (r , s) = D GL (|r r | , s) d + GL (|ri r | , s)0 (r , t0 )d, n

60


where r and r , ri is a position vector of i th point source.rr |,s) It is apparent that the potential kernels GL (|r r | , s) and GL (|n may be written more explicitly in the form: 1 GL (|r r | , s) = K0 (k R, s) (1.109) 2D ( ) GL (|r r | , s) 1 R k = K0 (k R, s) = K1 (k R, s) (1.110) n n 2D 2D n The derivative of the distance R = |r r | = (x x)2 + (y y )2 with respect to the unit outward normal n is calculated as for the steady state (see Eq.(1.91), Eq.(1.92) and Eq.(1.93)).

Therefore, the normal derivative of the Greens function can be rewritten in more explicit form as follows: [ ]( ) GL (r, s) k x x y y = K1 (k R, s) nx + ny (1.111) n 2D R R So the equation (1.108), under assumption that initial conditions of are equal to zero, can take the following form: ]( ) [ x x k y y c(r)(r, s) + K1 (k R, s) nx + ny (r , s)d = 2 R R (1.112) 1 (r , s) K0 (k R, s) d = n 2 The BIE of Eq.(1.108) can be solved numerically by dividing the boundary into elements and the domain into cells using appropriate shape functions. The numerical implementation is very similar to that of the steadystate problems described in Laplace equation section, since the form of singularity is of the same order. +1 M 1 2 G(|r r | , s) (j ) Nl ( )J ( )d = c(r)i (r, s) + l (r , s) n 1 j =0 l=0M 1 2 j =0 l=0 (j ) l (r , s)

(1.113)+1

=

n

1

1 G(|r r | , s)Nl ( )J ( )d 2


61

where, for simplicity we have skipped subscript L understanding that notation G(|r r | , s) means the Laplaces transform of fundamental solution, M is the total number of elements and the index l expresses the local numbers of the nodes of boundary elements, in this case the isoparametric quadratic boundary elements. If one makes the assumption that is varying harmonically in time then Eq. (1.106) becomes: 1 (r, ) = 0 (r, t0 ) D D

2 (r, ) i

(1.114)

where i =

1, = 2f is an angular frequency, k 2 = and k = i . D D

In this case the wave number is complex, so we have a modied Bessel function of complex arguments.

1.4.1

Treatment of singularity

In case of the solution in the frequency domain the singular integrals are treated in the same way as in case of the steady state (see for example section 1.2.6).

1.4.2


For the frequency domain formulation, the internal point values like or [ ]T , can be calculated in the similar way as it was described in section x y 1.3.2 (see Eqs. (1.101) and (1.102)).

62


1.5

Examples

To show the exibility and powerfulness of the BEM let consider two examples described by the Helmholtz equation. The rst one is for the Cartesian coordinate system, while the second one is for the polar coordinate one.

1.5.1

Cartesian coordinate system

To start with, let consider the Helmholtz equation in Cartesian coordinate system of the form [49]. 2 H = i H (1.115)

where i = 1 and [ s1 ] is the angular frequency, [1/( m)] is a material conductivity and [H/m] is a magnetic permeability. Let consider the following example: the onedimensional bounded conducting plate of thickness 2 b is placed in the sinusoidal magnetic eld (see Fig. 1.22).

z0 1 0 1 11111111111 00000000000 0 1 0 1 00000000000 11111111111 0 1 0 1 00000000000 11111111111 0 1 0 1 00000000000 11111111111 0 1 0 1 00000000000 11111111111 H z 1z 0 1 0 1 00000000000 11111111111 0 1 0 1 00000000000 11111111111 0 1 0 1 00000000000 11111111111 0 1 b b 0 1 00000000000 11111111111 0 1 0 1 00000000000 11111111111 0 1 0 1 00000000000 11111111111 0 1 J x 1x 11111111111 0 1 00000000000 0 1 0 1 00000000000 11111111111 0 1 0 1 00000000000 11111111111 0 1 0 1 00000000000 11111111111 0 1 0 1 00000000000 11111111111 0 1 x 0 1 00000000000 11111111111 0 1 0 1 00000000000 11111111111 0 1 0 1 00000000000 11111111111 0 1 00000000000 11111111111

y

Figure 1.22: Conducting plate of nite thickness

Both magnetic eld strength and eddy currents have one component only, in z and x direction, respectively. Solving Eq.(1.115) the analytical solution for

1.5. EXAMPLES eddy current component in x direction takes the form: Jx (y ) = where = dHz sinh(y) = Hz b dy cosh( ) 2

63

(1.116)

i and b is a thickness of the plate.

Numerical solution, in the frequency domain of Eq.(1.115) for two dierent boundary discretization, is presented in the Fig. 1.23 and Fig. 1.24. The solid line is the analytical solution, but the diamonds represents the boundary element solution. The relative error for the modulus and the angle32 boundary elements 5 4.5 4 3.5 3 2.5 2 1.5 1 0.5 0 0.05 0 0.05100 150 0.05 0 0.05 0 50 100 50 32 boundary elements 150 analitical solution BEM solution

analitical solution BEM solution

Figure 1.23: Modulus (left) and phase shift (right) of the function Hz changes across the plate thickness for 32 boundary elements discretization

128 boundary elements 5 4.5 4 3.5 3 2.5 2 1.5 1 0.5 0 0.05 0 0.05 100 150 0.05 0 50 100 50 analitical solution BEM solution 150

128 boundary elements analitical solution BEM solution

0

0.05

Figure 1.24: Modulus (left) and phase shift (right) of the function Hz changes across the plate thickness for 128 boundary elements discretization

64


shift for the second order boundary element are presented in the following Fig. 1.25.55 50 45 40 35 30 25 20 15 10 5 0 16 32 64 96 128 256 55 50 45 40 35 30 25 20 15 10 5 0 16 32 64 96 128 256

Figure 1.25: Maximal relative error for modulus (left) and maximal relative error for the phase shifting (right) in function of number of boundary elements

As we can see from the Fig. 1.25 the phase shift is much more sensitive than the magnitude and demands better discretization in order to achieve the same relative error. This example shows that the BE method is very ecient because even for the coarse discretization the relative error for both the magnitude and the phase shift is really small (see for example discretization for 96 boundary elements for which the maximal relative error is less than 5%).

1.5.2

Polar coordinate system

Now, let consider more complicated example described by the 2D Helmholtz equation on a disc radius r = b [m], which we express in polar coordinates [7]: 2 (r, ) + k 2 (r, ) = 0 where k is a wave number. For this equation it exists an analytical solution of the form: (r, ) = n=0

(1.117)

cn In (k r)ein

(1.118)

1.5. EXAMPLES

65

where In is the modied Bessel function of the rst kind and the n th order. The state function and its normal derivative on the boundary are then n given by: (r, )|r=b = n=0

cn In (k b)ein (1.119)

(r, ) n

=r =b

n=0

cn k In (k b)ein

Assume that the Robin boundary condition applies, such that a known input ux is specied leading to a boundary condition = D ( ) on the right hand side of: (b, ) + 2A D (b, ) = (b, ), n b (1.120)

where: D is the diusion coecient and coecient A depends on refractive index (see chapter 2.9.1). Than we have: n=0 ( ) cn In (k b) + 2A Dk In (k b) ein = n ein n=0

(1.121)

which leads to cn = n . In (k b) + 2A Dk In (k b) (1.122)

Now we have the analytic solution expressed as an innite series of modied Bessel functions.

1.5.3

Distributed source for a diusion model for light transport

Using this simple example (simple in the sense that this example posses analytic solution, see Eq.(1.119)), we consider three cases for the distributed

66


source on the boundary, when in Eq.(1.121), we will take n = 0, n = 3 and n = 9 terms of the solution. For n = 0 the Robin boundary condition will take the form: ( ) 0 = c0 I0 (k b) + 2A Dk I0 (k b) .

(1.123)

It means that incoming ux is constant regarding the magnitude and the phase shift along the whole boundary. That is why in this case only the distribution of the and the along the radius of the disc are presented n (see Fig. 1.26). For the boundary conditions dened by the Eq.(1.123), even coarse discretization (48 elements of the second order) produces nice results an relative error less than 5% (see Fig. 1.27). When the number of boundary elements will be increased to 192 then the relative error will drop below 1%. For n = 3 the Robin boundary condition will take the following form: ( ) 3 = c3 I3 (k b) + 2A Dk I3 (k b) ,

(1.124)

where I3 is the modied Bessel function of the rst kind and the third order. The distribution of the and n

along the boundary is presented in Fig. 1.28.

For this kind of boundary conditions we need more boundary elements in order to keep the relative error on the same level as in case of n = 0. For n = 9 the Robin boundary condition becomes: ( ) 9 = c9 I9 (k b) + 2ADk I9 (k b) .

(1.125)

where I9 is the modied Bessel function of the rst kind and the ninth order. The distribution of the state function and its normal derivative along the perimeter of the region is presented in the Fig. 1.30.

1.5. EXAMPLES

67

Figure 1.26: Analytical solution (solid lines) and numerical one for 48 boundary elements and n = 0 versus the radius of the disc4.5 4 3.5 relative error [%] relative error [%] 3 2.5 2 1.5 1 0.5 0 48 96 192 number of elements 4.5 4 3.5 3 2.5 2 1.5 1 0.5 0 48 96 192 number of elements

Figure 1.27: The modulus of the (left) and the phase shift of the (right) as a function of the boundary elements number

68


Figure 1.28: Analytical solution for n = 3 along the perimeter of the disc in radians

6 5 relative error [%] 4 3 2 1 0 48 96 192 384 768 number of the second order boundary elements

Figure 1.29: The modulus of the as a function of discretization density

The numerical solution for 96 second order boundary elements is shown in the next gure (see Fig. 1.31). In this case the oscillation of the external

1.5. EXAMPLES

69

Figure 1.30: Analytical solution for n = 9 along the perimeter of the disc in radians

values of and are clearly visible. The relative error distribution with n respect to the number of boundary elements is presented in the Fig. 1.32.

Concluding this relatively simple example when we have to deal with a distributed source, which is modelled by the Robin boundary conditions, the precision of the solution strongly depends on kind

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