finite element method for magnetic field response … · receiver on the ground surface. the finite...

FINITE ELEMENT METHOD FOR MAGNETIC FIELD RESPONSE FROM EXPONENTIAL CONDUCTIVITIES GROUND PROFILE

By Priyanuch Tunnurak

A Thesis Submitted in Partial Fulfillment of the Requirements for the Degree Master of Science Program in Mathematics

Department of Mathematics Graduate School, Silpakorn University

Academic Year 2014 Copyright of Graduate School, Silpakorn University

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สมดกลาง

FINITE ELEMENT METHOD FOR MAGNETIC FIELD RESPONSE FROM EXPONENTIAL CONDUCTIVITIES GROUND PROFILE

By

Priyanuch Tunnurak

A Thesis Submitted in Partial Fulfillment of the Requirements for the Degree

Master of Science Program in Mathematics

Department of Mathematics

Graduate School, Silpakorn University

Academic Year 2014

Copyright of Graduate School, Silpakorn University

สำนกหอ


วธไฟไนตเอลเมนตสาหรบสนามแมเหลกทตอบสนองมาจากพนดนทมสภาพนาไฟฟาแปรเปลยนแบบเอกซโพแนนเชยล

โดย นางสาวปรยานช ตนนรกษ

วทยานพนธนเปนสวนหนงของการศกษาตามหลกสตรปรญญาวทยาศาสตรมหาบณฑต สาขาวชาคณตศาสตร ภาควชาคณตศาสตร

บณฑตวทยาลย มหาวทยาลยศลปากร ปการศกษา 2557

ลขสทธของบณฑตวทยาลย มหาวทยาลยศลปากร

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The Graduate School, Silpakorn University has approved and accredited the Thesis title of “Finite Element Method for Magnetic Field Response from Exponential Conductivities Ground Profile” submitted by Miss Priyanuch Tunnurak as a partial fulfillment of the requirements for the degree of Master of Science in Mathematics

…...............................................................................

(Associate Professor Panjai Tantatsanawong, Ph.D.)

Dean of Graduate School

........../..................../..........

The Thesis Advisor Associate Professor Suabsagun Yooyuanyong, Ph.D. The Thesis Examination Committee .................................................... Chairman

(Assistant Professor Klot Patanarapeelert, Ph.D.)

............/......................../..............

.................................................... Member

(Warin Sripanya, Ph.D.)

............/......................../..............

.................................................... Member

(Associate Professor Suabsagun Yooyuanyong, Ph.D.)

............/......................../..............

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54305202 : MAJOR : MATHEMATICS KEY WORDS : FINITE ELEMENT / MAGNETIC / GALERKIN PRIYANUCH TUNNURAK : FINITE ELEMENT METHOD FOR MAGNETIC FIELD RESPONSE FROM EXPONENTIAL CONDUCTIVITIES GROUND PROFILE. THESIS ADVISOR : ASSOC. PROF. SUABSAGUN YOOYUANYONG, Ph.D. 182 pp. In this thesis, mathematical models of the magnetic field with two types of exponential conductivity are presented. The purpose of the models is to find the magnetic field at various locations under the condition that the Earth structure having exponential conductivities profile. There is a source providing a DC voltage and a receiver on the ground surface. The finite element method (FEM) is introduced by using the Galerkin's method of Weighted Residuals to find approximate solutions of partial differential equation. The numerical results are conducted to calculate and plot graphs of magnetic field at various locations. The results perform very well to the intensity of magnetic field for cross sections of ground structure. Department of Mathematics Graduate School, Silpakorn University

Student's signature ........................................ Academic Year 2014

Thesis Advisor's signature ........................................

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54305202 : สาขาวชาคณตศาสตร

คาสาคญ : ไฟไนตเอลเมนต / แมเหลก / กาเลอรคน

ปรยานช ตนนรกษ : วธไฟไนตเอลเมนตสาหรบสนามแมเหลกทตอบสนองมาจากพนดนทมสภาพนาไฟฟาแปรเปลยนแบบเอกซโพแนนเชยล. อาจารยทปรกษาวทยานพนธ : รศ.ดร.สบสกล อยยนยง. 182 หนา. ในวทยานพนธน เรานาเสนอแบบจาลองคณตศาสตรของสนามแมเหลกทตอบสนองจากพนดนทมสภาพนาไฟฟาแบบฟงกชนเอกซโพแนนเชยล 2 ชนด โดยใชวธไฟไนตเอลเมนตเพอหาสนามแมเหลกทตาแหนงตางๆ เราสมมตวาโครงสรางของพนโลกมชนเดยว โดยสภาพนาไฟฟามลกษณะการเปลยนแปลงแบบฟงกชนเอกซโพแนนเชยล มแหลงกาเนดไฟฟากระแสตรงและเครองรบสญญาณอยบนพนผวโลก เราใชวธเวทเรซดวของกาเลอรคนรวมกบวธไฟไนตเอลเมนตในการหาผลเฉลยโดยประมาณของสมการอนพนธยอย และใชโปรแกรม MATLAB ในการคานวณและ plot กราฟสนามแมเหลกทตาแหนงตางๆ ภาควชาคณตศาสตร บณฑตวทยาลย มหาวทยาลยศลปากร

ลายมอชอนกศกษา........................................ ปการศกษา 2557

ลายมอชออาจารยทปรกษาวทยานพนธ ........................................

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Acknowledgements

This thesis, I would like to express my gratitude and appreciation to my thesis advisor, Assoc. Prof. Dr. Suabsagun Yooyuanyong, for his expertise, understanding and patience. I appreciate his knowledge, skills in many areas and his assistance in writing reports. I would like to thank Dr. Warin Sripanya and Mr. Preecha Lee for their helpful suggestions, advice and support. I would like to give a very special thanks to Dr. Nairat Kanyamee for her expertise, understanding and patience. I appreciate her knowledge and assistance in writing reports. The most important, I would like to express the gratitude from my heart to my beloved parents and my relatives for love, understanding and continuous support. I would like to thank my friends for the support and encouragement whenever I was in need. Last but not least, I would like to thank the Department of Mathematics, Faculty of Science, Silpakorn University and Centre of Excellence in Mathematics for continuous support.

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Table of Contents

Page

Abstract in English .................................................................................................. d Abstract in Thai ....................................................................................................... e Acknowledgments................................................................................................... f List of Figures ......................................................................................................... h List of Tables .......................................................................................................... m Chapter 1 Introduction ................................................................................................. 1 2 Galerkin’s Method of Weighted Residual ................................................... 4 3 Numerical formulations ............................................................................... 12 4 Numerical Results for the Case of σ(z) = σ0e-bz .......................................... 22 5 Numerical Results for the Case of σ(z) = σ0+(σ1-σ0)e-bz ............................. 88 6 Results and Discussion ................................................................................ 148 References ............................................................................................................... 150 Appendix ................................................................................................................. 154 Biography ................................................................................................................ 182

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List of Figures

Figures Page

2.1 A finite element rectangulation ............................................................ 8 2.2 The basis function ϕj ............................................................................ 9 2.3 Local coordinate transformation .......................................................... 10 3.1 Geometric model of the Earth structure ............................................... 12 3.2 Boundary conditions of the Earth structure ......................................... 16 3.3 The coordinate transformation (r, z) in terms of the local coordinates (ξ, η) ..................................................................................................... 18 3.4 The nodes of the elements ........................................................ 19 3.5 The basis function ϕj ............................................................................ 19 4.1 Geometric model of the Earth structure in the case of σ(z) = σ0e-bz .... 22 Case of an exponentially decreasing conductivity σ(z) = σ0e-bz 4.2 The value of magnetic field when b = 0.001 m-1 ................................. 40 4.3 The relationship between magnetic field and distance of receiver From source at various depths as b = 0.001 m-1. (a) The value of magnetic field when 10 ≤ r ≤ 170 m and 0 ≤ z ≤ 160 m. (b) The value of magnetic field when z is fixed, i.e. z = 0, 20, 40, . . . , 160 m (from top to bottom) ............................. 41 4.4 The value of magnetic field when b = 0.01 m-1 ................................... 43 4.5 The relationship between magnetic field and distance of receiver From source at various depths as b = 0.01 m-1. (a) The value of magnetic field when 10 ≤ r ≤ 170 m and 0 ≤ z ≤ 160 m. (b) The value of magnetic field when z is fixed, i.e. z = 0, 20, 40, . . . , 160 m (from top to bottom) ............................. 44 4.6 The value of magnetic field when b = 0.05 m-1 ................................... 46 4.7 The relationship between magnetic field and distance of receiver From source at various depths as b = 0.05 m-1. (a) The value of magnetic field when 10 ≤ r ≤ 170 m and 0 ≤ z ≤ 160 m. (b) The value of magnetic field when z is fixed, i.e. z = 0, 20, 40, . . . , 160 m (from top to bottom) ............................. 47 4.8 The value of magnetic field when b = 0.075 m-1 ................................. 49 4.9 The relationship between magnetic field and distance of receiver From source at various depths as b = 0.075 m-1. (a) The value of magnetic field when 10 ≤ r ≤ 170 m and 0 ≤ z ≤ 160 m. (b) The value of magnetic field when z is fixed, i.e. z = 0, 20, 40, . . . , 160 m (from top to bottom) ............................. 50 4.10 The value of magnetic field when b = 0.1 m-1 ..................................... 52

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4.11 The relationship between magnetic field and distance of receiver From source at various depths as b = 0.1 m-1. (a) The value of magnetic field when 10 ≤ r ≤ 170 m and 0 ≤ z ≤ 160 m. (b) The value of magnetic field when z is fixed, i.e. z = 0, 20, 40, . . . , 160 m (from top to bottom) ............................. 53 4.12 The relationship between magnetic field and distance of receiver from source at various depths when (a) b = 0.001 m-1 (b) b = 0.01 m-1 (c) b = 0.05 m-1 (d) b = 0.075 m-1 (e) b = 0.1 m-1 ................................. 55 4.13 The relationship between magnetic field and distance of receiver from source when z is fixed and (a) b = 0.001 m-1 (b) b = 0.01 m-1 (c) b = 0.05 m-1 (d) b = 0.075 m-1 (e) b = 0.1 m-1 ................................. 57 4.14 The relationship between magnetic field and different depths when r is fixed and (a) b = 0.001 m-1 (b) b = 0.01 m-1 (c) b = 0.05 m-1 (d) b = 0.075 m-1 (e) b = 0.1 m-1 ........................................................... 59 4.15 The relationship between magnetic field and distance of receiver from source when b varies from 0.001, 0.01, 0.05, 0.075 and 0.1 m-1 and z is fixed. (a) z=20 m (b) z=60 m (c) z=100 m (d) z=140 m .......... 61 4.16 Contour graphs of magnetic field at different distances of receiver

from source and different depths when (a) b = 0.001 m-1 (b) b = 0.01 m-1 (c) b = 0.05 m-1 (d) b = 0.075 m-1 (e) b = 0.1 m-1 ....... 62 Case of an exponentially Increasing conductivity σ(z) = σ0e-bz 4.17 The value of magnetic field when b = 0.001 m-1 ................................. 64 4.18 The relationship between magnetic field and distance of receiver From source at various depths as b = 0.001 m-1. (a) The value of magnetic field when 10 ≤ r ≤ 170 m and 0 ≤ z ≤ 160 m. (b) The value of magnetic field when z is fixed, i.e. z = 0, 20, 40, . . . , 160 m (from top to bottom) ............................. 65 4.19 The value of magnetic field when b = 0.01 m-1 ................................... 67 4.20 The relationship between magnetic field and distance of receiver From source at various depths as b = 0.01 m-1. (a) The value of magnetic field when 10 ≤ r ≤ 170 m and 0 ≤ z ≤ 160 m. (b) The value of magnetic field when z is fixed, i.e. z = 0, 20, 40, . . . , 160 m (from top to bottom) ............................. 68 4.21 The value of magnetic field when b = 0.05 m-1 ................................... 70 4.22 The relationship between magnetic field and distance of receiver From source at various depths as b = 0.05 m-1. (a) The value of magnetic field when 10 ≤ r ≤ 170 m and 0 ≤ z ≤ 160 m. (b) The value of magnetic field when z is fixed, i.e. z = 0, 20, 40, . . . , 160 m (from top to bottom) ............................. 71 4.23 The value of magnetic field when b = 0.075 m-1 ................................. 73

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4.24 The relationship between magnetic field and distance of receiver From source at various depths as b = 0.075 m-1. (a) The value of magnetic field when 10 ≤ r ≤ 170 m and 0 ≤ z ≤ 160 m. (b) The value of magnetic field when z is fixed, i.e. z = 0, 20, 40, . . . , 160 m (from top to bottom) ............................. 74 4.25 The value of magnetic field when b = 0.1 m-1 ..................................... 76 4.26 The relationship between magnetic field and distance of receiver From source at various depths as b = 0.1 m-1. (a) The value of magnetic field when 10 ≤ r ≤ 170 m and 0 ≤ z ≤ 160 m. (b) The value of magnetic field when z is fixed, i.e. z = 0, 20, 40, . . . , 160 m (from top to bottom) ............................. 77 4.27 The relationship between magnetic field and distance of receiver from source at various depths when (a) b = 0.001 m-1 (b) b = 0.01 m-1 (c) b = 0.05 m-1 (d) b = 0.075 m-1 (e) b = 0.1 m-1 ................................. 79 4.28 The relationship between magnetic field and distance of receiver from source when z is fixed and (a) b = 0.001 m-1 (b) b = 0.01 m-1 (c) b = 0.05 m-1 (d) b = 0.075 m-1 (e) b = 0.1 m-1 ................................. 81 4.29 The relationship between magnetic field and different depths when r is fixed and (a) b = 0.001 m-1 (b) b = 0.01 m-1 (c) b = 0.05 m-1 (d) b = 0.075 m-1 (e) b = 0.1 m-1 ........................................................... 83 4.30 The relationship between magnetic field and distance of receiver from source when b varies from 0.001, 0.01, 0.05, 0.075 and 0.1 m-1 and z is fixed. (a) z=20 m (b) z=60 m (c) z=100 m (d) z=140 m .......... 85 4.31 Contour graphs of magnetic field at different distances of receiver

from source and different depths when (a) b = 0.001 m-1 (b) b = 0.01 m-1 (c) b = 0.05 m-1 (d) b = 0.075 m-1 (e) b = 0.1 m-1 ....... 86 5.1 Geometric model of the Earth structure in the case of σ(z) = σ0+(σ1-σ0)e-bz ............................................................................. 88 Case of an exponentially decreasing conductivity σ0+(σ1-σ0)e-bz 5.2 The value of magnetic field when b = 0.01 m-1 ................................... 100 5.3 The relationship between magnetic field and distance of receiver From source at various depths as b = 0.01 m-1. (a) The value of magnetic field when 10 ≤ r ≤ 170 m and 0 ≤ z ≤ 160 m. (b) The value of magnetic field when z is fixed, i.e. z = 0, 20, 40, . . . , 160 m (from top to bottom) ............................. 101 5.4 The value of magnetic field when b = 0.05 m-1 ................................... 103 5.5 The relationship between magnetic field and distance of receiver From source at various depths as b = 0.05 m-1. (a) The value of magnetic field when 10 ≤ r ≤ 170 m and 0 ≤ z ≤ 160 m. (b) The value of magnetic field when z is fixed, i.e. z = 0, 20, 40, . . . , 160 m (from top to bottom) ............................. 104

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5.6 The value of magnetic field when b = 0.1 m-1 ..................................... 106 5.7 The relationship between magnetic field and distance of receiver From source at various depths as b = 0.1 m-1. (a) The value of magnetic field when 10 ≤ r ≤ 170 m and 0 ≤ z ≤ 160 m. (b) The value of magnetic field when z is fixed, i.e. z = 0, 20, 40, . . . , 160 m (from top to bottom) ............................. 107 5.8 The value of magnetic field when b = 0.2 m-1 ..................................... 109 5.9 The relationship between magnetic field and distance of receiver From source at various depths as b = 0.2 m-1. (a) The value of magnetic field when 10 ≤ r ≤ 170 m and 0 ≤ z ≤ 160 m. (b) The value of magnetic field when z is fixed, i.e. z = 0, 20, 40, . . . , 160 m (from top to bottom) ............................. 110 5.10 The value of magnetic field when b = 0.3 m-1 ..................................... 112 5.11 The relationship between magnetic field and distance of receiver From source at various depths as b = 0.3 m-1. (a) The value of magnetic field when 10 ≤ r ≤ 170 m and 0 ≤ z ≤ 160 m. (b) The value of magnetic field when z is fixed, i.e. z = 0, 20, 40, . . . , 160 m (from top to bottom) ............................. 113 5.12 The relationship between magnetic field and distance of receiver from source at various depths when (a) b = 0.01 m-1 (b) b = 0.05 m-1 (c) b = 0.1 m-1 (d) b = 0.2 m-1 (e) b = 0.3 m-1 ....................................... 115 5.13 The relationship between magnetic field and distance of receiver from source when z is fixed and (a) b = 0.01 m-1 (b) b = 0.05 m-1 (c) b = 0.1 m-1 (d) b = 0.2 m-1 (e) b = 0.3 m-1 ....................................... 117 5.14 The relationship between magnetic field and different depths when r is fixed and (a) b = 0.01 m-1 (b) b = 0.05 m-1 (c) b = 0.1 m-1 (d) b = 0.2 m-1 (e) b = 0.3 m-1 ............................................................... 119 5.15 The relationship between magnetic field and distance of receiver from source when b varies from 0.01, 0.05, 0.1, 0.2 and 0.3 m-1 and z is fixed. (a) z=20 m (b) z=60 m (c) z=100 m (d) z=140 m .......... 121 5.16 Contour graphs of magnetic field at different distances of receiver

from source and different depths when (a) b = 0.01 m-1 (b) b = 0.05 m-1 (c) b = 0.1 m-1 (d) b = 0.2 m-1 (e) b = 0.3 m-1 ............. 122 Case of an exponentially Increasing conductivity σ0+(σ1-σ0)e-bz 5.17 The value of magnetic field when b = 0.01 m-1 ................................... 124 5.18 The relationship between magnetic field and distance of receiver From source at various depths as b = 0.01 m-1. (a) The value of magnetic field when 10 ≤ r ≤ 170 m and 0 ≤ z ≤ 160 m. (b) The value of magnetic field when z is fixed, i.e. z = 0, 20, 40, . . . , 160 m (from top to bottom) ............................. 125 5.19 The value of magnetic field when b = 0.05 m-1 ................................... 127

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5.20 The relationship between magnetic field and distance of receiver From source at various depths as b = 0.05 m-1. (a) The value of magnetic field when 10 ≤ r ≤ 170 m and 0 ≤ z ≤ 160 m. (b) The value of magnetic field when z is fixed, i.e. z = 0, 20, 40, . . . , 160 m (from top to bottom) ............................. 128 5.21 The value of magnetic field when b = 0.1 m-1 ..................................... 130 5.22 The relationship between magnetic field and distance of receiver From source at various depths as b = 0.1 m-1. (a) The value of magnetic field when 10 ≤ r ≤ 170 m and 0 ≤ z ≤ 160 m. (b) The value of magnetic field when z is fixed, i.e. z = 0, 20, 40, . . . , 160 m (from top to bottom) ............................. 131 5.23 The value of magnetic field when b = 0.2 m-1 ..................................... 133 5.24 The relationship between magnetic field and distance of receiver From source at various depths as b = 0.2 m-1. (a) The value of magnetic field when 10 ≤ r ≤ 170 m and 0 ≤ z ≤ 160 m. (b) The value of magnetic field when z is fixed, i.e. z = 0, 20, 40, . . . , 160 m (from top to bottom) ............................. 134 5.25 The value of magnetic field when b = 0.3 m-1 ..................................... 136 5.26 The relationship between magnetic field and distance of receiver From source at various depths as b = 0.3 m-1. (a) The value of magnetic field when 10 ≤ r ≤ 170 m and 0 ≤ z ≤ 160 m. (b) The value of magnetic field when z is fixed, i.e. z = 0, 20, 40, . . . , 160 m (from top to bottom) ............................. 137 5.27 The relationship between magnetic field and distance of receiver from source at various depths when (a) b = 0.01 m-1 (b) b = 0.05 m-1 (c) b = 0.1 m-1 (d) b = 0.2 m-1 (e) b = 0.3 m-1 ....................................... 139 5.28 The relationship between magnetic field and distance of receiver from source when z is fixed and (a) b = 0.01 m-1 (b) b = 0.05 m-1 (c) b = 0.1 m-1 (d) b = 0.2 m-1 (e) b = 0.3 m-1 ....................................... 141 5.29 The relationship between magnetic field and different depths when r is fixed and (a) b = 0.01 m-1 (b) b = 0.05 m-1 (c) b = 0.1 m-1 (d) b = 0.2 m-1 (e) b = 0.3 m-1 ............................................................... 143 5.30 The relationship between magnetic field and distance of receiver from source when b varies from 0.01, 0.05, 0.1, 0.2 and 0.3 m-1 and z is fixed. (a) z=20 m (b) z=60 m (c) z=100 m (d) z=140 m .......... 145 5.31 Contour graphs of magnetic field at different distances of receiver

from source and different depths when (a) b = 0.01 m-1 (b) b = 0.05 m-1 (c) b = 0.1 m-1 (d) b = 0.2 m-1 (e) b = 0.3 m-1 ............. 146

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List of Tables

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4.1 The value of in

coordinates (ξ, η) .................................................................................. 24

4.2 The value of in coordinates

(ξ, η) ..................................................................................................... 30

4.3 The value of in coordinates (ξ, η) ............ 34

5.1 The value of

in

coordinates (ξ, η) .................................................................................. 90

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Chapter 1

Introduction

Currently, human studied the Earth structure widely in order to utilize

the natural resources embedded beneath the Earth surface for developing the

agricultural sector and industrial sectors in their countries. They use knowledge of

geophysics which is a branch of science. The survey uses mathematics, physics and

the physical properties of the Earth such as the resistivity, conductivity, electric

potential, magnetic field and electric field to search for the natural resources under

the ground.

Geophysics is divided into two great divisions, the global geophysics and

exploration geophysics. For the global geophysics, we study the earthquakes, phys-

ical oceanography, the Earth thermal state and meteorology. For the exploration

geophysics, we study the physical properties of the Earth then apply them in the

search for oil, gas, gold and minerals embedded beneath the Earth [37].

We use the exploration geophysics to search for the natural resources be-

neath the Earth surface to differentiate the minerals from the others and process

the data obtained from a geophysics survey to identify the location of minerals

correctly. Since the most natural resources embedded beneath the Earth is hard to

find. This geophysics survey can be costly, so we seek for a mathematical model

which is a method that became famous because it is economical and costs less

than the direct survey.

The survey method of the geological structure of the Earth by using the

knowledge of geophysics contains electromagnetic method, resistivity method and

magnetometric resistivity method. We use the magnetometric resistivity method

to survey the natural resources because it uses a low-frequency magnetic field and

1

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is not induced by the electricity in the Earth. As a result, it is more accurate than

the other survey methods [10].

We create a mathematical model by using magnetometric resistivity method

to find the value of magnetic field beneath the Earth surface. In 2003, Chen and

Oldenburg [7] assume that the Earth structure consists of horizontally stratified

layers having constant conductivities at certain depths except the last layer where

the conductivity having the same varying through the rest of the layer. They

derived the magnetic field directly by solving a boundary value problem of a

horizontally stratified layered Earth with homogeneous layers. However, in the

real situation there are some cases where the subsurface conductivities vary ex-

ponentially, linearly or binomially with depth, such as coastal areas and areas

near water sources. There exists a considerable amount of research about mathe-

matical modeling which assumes that the Earth structure consists of horizontally

stratified multilayer with one or more layers having exponentially, linearly or bi-

nomially varying conductivities at certain depths except the last layer where the

conductivity having the same varying through the rest of the layer. Stoyer and

Wait [28] studied the problem of computing apparent resistivity for a structure

with a homogeneous overburden overlying a medium whose resistivity varies expo-

nentially with depth. Banerjee et al. [1] gave expressions for apparent resistivity

of a multilayered Earth with a layer having exponentially varying conductivity.

Kim and Lee [14] derived a new resistivity kernel function for calculating appar-

ent resistivity of a multilayered Earth with layers having exponentially varying

conductivities. Siew and Yooyuanyong [29] studied the electromagnetic response

of a thin disk beneath an inhomogeneous conductive overburden. They derived ex-

pressions for the electric fields above the ground surface. Ketchanwit [15] studied

the Earth surface layers using time-domain electromagnetic field by constructing

mathematical models of the ground having exponentially varying and constant

varying conductivities. Sripunya [30] derived solutions of the steady state mag-

netic field due to a DC current source in a layered Earth with some layer having

exponentially or binomially or linearly varying conductivity.

In this thesis, the mathematical model is proposed by using numerical tech-

niques for finding approximate solutions. The finite element method (FEM) is used

to find the numerical solutions of the magnetic field under the Earth surface. We

assume two kinds of geophysics models that the Earth structure contains only one

layer having exponential conductivities(σ = σ0e−bz and σ(z) = σ0+(σ1−σ0)e

−bz).

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This method that proposed in this thesis is different from the Hankel transform

approach which is difficult to solve for some complex problems such as all the

research articles mentioned above. There are a few research using the Finite El-

ement Method (FEM) by applying the Galerkin’s method of Weighted Residuals

to find the solution of the magnetic field. Lee [16] presented a numerical method

to compute the electromagnetic response of two-dimensional Earth models. Ve-

limsky and Martince [36] introduced a time-domain method to solve the prob-

lem of geomagnetic induction in a heterogeneous Earth excited by variations of

the ionospheric and magnetospheric currents with arbitrary spatiotemporal char-

acteristics. Mitsuhata and Uchida [20] presented a finite element algorithm for

computing magnetic field response for 3D conductivity structures. Therefore, we

are interested in approximation techniques in finding the magnetic field beneath

the Earth by using the Galerkin’s method of Weighted Residuals. ส

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Chapter 2

Galerkin’s Method of Weighted

Residuals

In this chapter, the Galerkin’s method of weighted residuals is emphasized

as a tool for finite element formulation for a problem governed by differential

equation from David Hutton [12].

2.1 Method of weighted residuals

The method of weighted residuals (MWR) is an approximate technique

for solving boundary value problems that utilizes trial functions satisfying the

prescribed boundary conditions and integral formulation to minimize error, in an

average sense, over the problem domain. The general concept is described here

in terms of the one-dimensional case. Given a differential equation of the general

form

D[y(x), x] = 0 a < x < b, (2.1)

where D is differential operator subject to homogeneous boundary conditions

y(a) = y(b) = 0, (2.2)

the method of weighted residuals seeks an approximate solution in the form

y∗(x) =n∑

i=1

ciNi(x), (2.3)

where y∗ is the approximate solution expressed as the product of ci unknown,

constant parameters to be determined and Ni(x) trial functions. The major re-

quirement placed on the trial functions is that they be admissible functions; that

4

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5

is, the trial functions are continuous over the domain of interest and satisfy the

specified boundary conditions exactly. In addition, the trial functions should be

selected to satisfy the ”physics” of the problem in a general sense. Given these

somewhat lax conditions, it is highly unlikely that the solution represented by

Equation (2.3) is exact. Instead, on substitution of the assumed solution into

the differential Equation (2.1), a residual error (hereafter simply called residual)

results such that

R(x) = D[y∗(x), x] �= 0, (2.4)

where R(x) is the residual. Note that the residual is also a function of the un-

known parameters ci. The method of weighted residuals requires that the unknown

parameters ci be evaluated such that∫ b

a

wi(x)R(x)dx = 0, i = 1, . . . , n, (2.5)

where wi(x) represents n arbitrary weighting functions. We observe that, on

integration, Equation (2.5) results in n algebraic equations, which can be solved

for the n values of ci. Equation (2.5) expresses that the sum (integral) of the

weighted residual error over the domain of the problem is zero. Owing to the

requirements placed on the trial functions, the solution is exact at the end points

(the boundary conditions must be satisfied) but, in general, at any interior point

the residual error is nonzero.

Several variations of MWR exist and the techniques vary primarily in how

the weighting factors are determined or selected. The most common techniques are

point collocation, subdomain collocation, least squares, and Galerkin’s method.

As it is quite simple to use and readily adaptable to the finite element method,

we discuss only Galerkin’s method.

In Galerkin’s weighted residual method, the weighting functions are chosen

to be identical to the trial functions; that is,

wi(x) = Ni(x), i = 1, . . . , n. (2.6)

Therefore, the unknown parameters are determined via∫ b

a

wi(x)R(x)dx =

∫ b

a

Ni(x)R(x)dx = 0, i = 1, . . . , n, (2.7)

again resulting in n algebraic equations for evaluation of the unknown parameters.

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6

2.2 The Galerkin finite element method

The classic method of weighted residuals described in the previous section

utilizes trial functions that are global; that is, each trial function must apply over

the entire domain of interest and identically satisfy the boundary conditions. Par-

ticularly in the more practical cases of two dimensional problems governed by

partial differential equations, ”discovery” of appropriate trial function and deter-

mination of the accuracy of the resulting solutions are formidable tasks. However,

the concept of minimizing the residual error is readily adapted to the finite ele-

ment context using the Galerkin approach as follows. For illustrative purposes,

we consider the Poisson equation

−Δu = f in Ω, (2.8a)

u = 0 on Γ, (2.8b)

where Ω is a bounded domain in the place with boundary Γ, f is a given real-valued

bounded function in Ω, and the Laplacian operator Δ is defined by

Δu =∂2u

∂x21

+∂2u

∂x22

. (2.9)

To write equation (2.8) in form weak formulation, we need to use Green’s formula,

which is an extension to multiple dimensions of integration∫Ω

ΔvwdX =

∫Γ

∂v

∂νwdl −

∫Ω

∇v · ∇wdX, (2.10)

where v, w ∈ V below, ν is the outward unit normal to Γ and the normal derivative

is expressed by∂v

∂ν=

∂v

∂x1

ν1 +∂v

∂x2

ν2.

We shall now give a weak formulation of problem (2.8) to find u ∈ H1. For

the two dimensional problem (2.8), the admissible function space V is composed

of the following real functions:

V = { v ∈ H1 : v is a continuous function on Ω, ∂v∂x1

and ∂v∂x2

are piecewise

continuous and bounded on Ω and v = 0 on Γ },where the Hilbert space

H1 = {v ∈ L2 :∂v

∂xi

∈ L2, i = 1, . . . , d},

L2 = {v : v is defined on Ω and

∫Ω

v2dX < ∞}.

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7

Multiplying the equation (2.8a) by v ∈ V and integrating over Ω, we see

that

(−Δu, v) = (f, v),

or

−∫Ω

ΔuvdX =

∫Ω

fvdX.

Applying Green’s formula (2.10) to this equation and using the homoge-

neous boundary condition in the space V lead to the weak formulation:

Find u ∈ H1 such that∫Ω

∇u · ∇vdX =

∫Ω

fvdX, ∀v ∈ V. (2.11)

We now construct a finite-dimensional subspace Vh ⊂ V . For simplicity,

we assume that Ω is a polygonal domain. We make a rectangulation of Ω, by

subdividing Ω into a set Kh = {K1, . . . , KM} of non-overlapping rectangles Ki,

Ω =⋃

K∈Kh

K = K1 ∪K2 . . . ∪KM ,

such that no vertex of one rectangle lies on the edge of another rectangle (see

Figure 2.1)

For rectangles K ∈ Kh, we define the mesh parameters

hK =diam(K) =the longest edge of K and h = maxK∈Kh

hK .

Figure 2.1: A finite element rectangulation.

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8

Now, we introduce the simplest finite element space in two dimensions

Vh = { v ∈ H1 : v is a continuous function on Ω, v is linear on each rectangle

K ∈ Kh and v = 0 on Γ }.

The finite element method for problem (2.8) is formulated as:

Find uh ∈ H1 such that∫Ω

∇uh · ∇vdX =

∫Ω

fvdX, ∀v ∈ Vh. (2.12)

As parameters to describe a function v ∈ Vh we choose the values v(Xi)

of v at the nodes Xi for each i = 1, . . . ,M of Kh (see Figure 2.1) but exclude

the nodes on the boundary since v = 0 on Γ. The corresponding basis functions

ϕj ∈ Vh for each j = 1, . . . ,M are then defined by (see Figure 2.2)

ϕj(Xi) =

{1 , i = j,

0 , i �= j, i, j = 1, . . . ,M.

Figure 2.2: The basis function ϕj.

The support of ϕj, i.e., the set of points X for which ϕj(X) �= 0, consists

of the rectangles with the common node Xj (Figure 2.2). The function ϕj is a

two-dimensional nets function. On each element K, these functions become the

local shape functions. Any function v ∈ Vh has the unique representation

v(X) =M∑j=1

αjϕj(X), X ∈ Ω,

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9

where αj = v(Xj). Due to the boundary condition imposed in the finite element

space Vh, we exclude the vertices on the boundary of Ω (i.e.,the boundary nodes).

In the same way as for equation (2.12), with

uh(X) =M∑j=1

ujϕj(X), uj = uh(Xj) X ∈ Ω. (2.13)

Equation (2.12) can be written in matrix form

Au = f,

where A = (ai,j), the stiffness matrix is an M × M matrix with elements ai,j =

a(ϕi, ϕj) =

∫Ω

∇ϕi · ∇ϕjdX and u = ui, f = fi are M-vectors with elements

ui = uh(Xi), fi = (f, ϕi) =

∫Ω

fϕidX, so equation (2.12) becomes

∫Ω

∇ϕi · ∇ϕjdX =

∫Ω

fϕidX, i, j = 1, . . . ,M. (2.14)

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10

Figure 2.3: Local coordinate transformation.

2.2.1 Local coordinate transformation

The local stiffness matrix A and right-hand side vector f can be obtained

using local coordinate system transformations. For a typical rectangle K in the

global coordinate system with vertices xi, xj, xm and xn, suppose that F is a one-

to-one mapping from the reference rectangle K in the ξη-coordinate system, with

vertices (−1,−1), (1,−1), (1, 1) and (−1, 1), onto K (Figure 2.3); these vertices

correspond to the node identifiers 1, 2, 3 and 4, respectively. The basis (shape)

functions ψj defined on the reference element K are

ψ1(ξ, η) =1

4(1− ξ)(1− η),

ψ2(ξ, η) =1

4(1 + ξ)(1− η),

ψ3(ξ, η) =1

4(1 + ξ)(1 + η),

ψ4(ξ, η) =1

4(1− ξ)(1 + η). (2.15)

Now, the transformation F is given by

X = F (ξ, η) = (x1(ξ, η), x2(ξ, η)). (2.16)

We suppose that the functions x1 and x2 are continuously differentiable

with respect to ξ and η. Then infinitesimals dξ and dη transform into dx1 and dx2

according to

dx1 =∂x1

∂ξdξ +

∂x1

∂ηdη and dx2 =

∂x2

∂ξdξ +

∂x2

∂ηdη,

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11

which can be written in matrix form as[dx1

dx2

]=

[∂x1

∂ξ∂x1

∂η∂x2

∂ξ∂x2

∂η

][dξ

dη

]. (2.17)

The 2× 2 matrix of partial derivatives in (2.17) is called the Jacobian matrix of

the transformation (2.16) and is denoted J .

A necessary and sufficient condition for the system (2.17) to be invertible

is that the determinant |J | of the Jacobian matrix be nonzero at (ξ, η) ∈ K. The

function |J | is called the Jacobian of the transformation (2.16),

|J | = det J =∂x1

∂ξ

∂x2

∂η− ∂x1

∂η

∂x2

∂ξ. (2.18)

The determinant of the Jacobian matrix is used for the transformation of

integrals from the global coordinate system to the local coordinate system:

dx1dx2 = |J |dξdη. (2.19)

Then we can convert ϕ to a function ϕ of ξ and η defined on K by setting

ϕ(x1, x2) = ϕ(x1(ξ, η), x2(ξ, η)) = ϕ(ξ, η).

Since equation (2.14) when the local coordinate transformation is used to obtain

the basis function, we obtain∫K

∇ϕi · ∇ϕjdX =

∫K

fϕidX, i, j = 1, . . . ,M. (2.20)

When we apply the change of variable F : K → K to equation (2.20), we have∫K

∇ϕi · ∇ϕj|J |dξdη =

∫K

fϕi|J |dξdη,

or ∫ 1

−1

∫ 1

−1

∇ϕi · ∇ϕj|J |dξdη =

∫ 1

−1

∫ 1

−1

fϕi|J |dξdη, i, j = 1, . . . ,M. (2.21)

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Chapter 3

Numerical formulations

In this chapter, we use finite element method (FEM) for constructing ap-

proximate solutions of problem. Assuming that the Earth structure contains only

one layer having exponential conductivity and there are a source providing a DC

voltage and a receiver on the ground surface which picks up the signal from r = 10

m to r = 190 m as shown in Figure 3.1.

Figure 3.1: Geometric model of the Earth structure.

We define z as the depth of an object from the Earth surface (meter), r as

the distance between source and receiver of magnetic field on the Earth surface

(meter) and σ(z) as the conductivity of the medium which is a function of z (S/m).

From Maxwell’s equations , the relationship between the electric and mag-

netic fields[29,30,31,37] can be written in cylindrical coordinates (r, φ, z) as follows

∇× �E = �0, (3.1)

12

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13

and

∇× �H = σ �E, (3.2)

where �E is the electric field vector, �H is the magnetic field vector, σ is the con-

ductivity of the medium and ∇ is the gradient operator in cylindrical coordinates

(r, φ, z) [17,27] defined by

∇ =∂

∂rer +

1

r

∂

∂φeφ +

∂

∂zez,

where er is the unit vector in radial direction (r), eφ is the unit vector in the

direction of φ, ez is the unit vector in the direction of z.

From (3.2) , we obtain

�E =1

σ(∇× �H). (3.3)

Substituting equation (3.3) into (3.1) , we obtain

∇× 1

σ(∇× �H) = �0. (3.4)

Let �H = Hrer +Hφeφ +Hzez.

The curl operator in cylindrical coordinates (r, φ, z) [17,27] is defined by

1

σ(∇× �H) =

1

r

∣∣∣∣∣∣∣∣er reφ ez

1σ

(∂∂r

)1σ

(∂∂φ

)1σ

(∂∂z

)Hr rHφ Hz

∣∣∣∣∣∣∣∣=

1

r

[(1

σ

∂Hz

∂φ− 1

σ

∂(rHφ)

∂z

)er +

(1

σ

∂Hr

∂z− 1

σ

∂Hz

∂r

)reφ

+

(1

σ

∂(rHφ)

∂r− 1

σ

∂Hr

∂φ

)ez

].

Then

∇× 1

σ(∇× �H) =

1

r

∣∣∣∣∣∣∣∣er reφ ez∂∂r

∂∂φ

∂∂z

1r

(1σ∂Hz

∂φ− 1

σ

∂(rHφ)

∂z

) (1σ∂Hr

∂z− 1

σ∂Hz

∂r

)1r

(1σ

∂(rHφ)

∂r− 1

σ∂Hr

∂φ

)∣∣∣∣∣∣∣∣

=1

r

[∂

∂φ

(1

r

(1

σ

∂(rHφ)

∂r− 1

σ

∂Hr

∂φ

))− ∂

∂z

(1

σ

∂Hr

∂z− 1

σ

∂Hz

∂r

)]er

+

[∂

∂z

(1

r

(1

σ

∂Hz

∂φ− 1

σ

∂(rHφ)

∂z

))− ∂

∂r

(1

r

(1

σ

∂(rHφ)

∂r− 1

σ

∂Hr

∂φ

))]eφ

+1

r

[∂

∂r

(1

σ

∂Hr

∂z− 1

σ

∂Hz

∂r

)− ∂

∂φ

(1

r

(1

σ

∂Hz

∂φ− 1

σ

∂(rHφ)

∂z

))]ez.

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14

Substitute equation ∇× 1σ(∇× �H) into (3.4) , we obtain

1

r

[∂

∂φ

(1

r

(1

σ

∂(rHφ)

∂r− 1

σ

∂Hr

∂φ

))− ∂

∂z

(1

σ

∂Hr

∂z− 1

σ

∂Hz

∂r

)]er

+

[∂

∂z

(1

r

(1

σ

∂Hz

∂φ− 1

σ

∂(rHφ)

∂z

))− ∂

∂r

(1

r

(1

σ

∂(rHφ)

∂r− 1

σ

∂Hr

∂φ

))]eφ

+1

r

[∂

∂r

(1

σ

∂Hr

∂z− 1

σ

∂Hz

∂r

)− ∂

∂φ

(1

r

(1

σ

∂Hz

∂φ− 1

σ

∂(rHφ)

∂z

))]ez = �0, (3.5)

where Hr, Hφ and Hz are the components of �H in er, eφ and ez directions, respec-

tively (i.e. �H = �H(r, φ, z)). Since the magnetic field is axisymmetric, it depends

only on r and z and not on the azimuth φ, i.e. �H = �H(r, z)[3,18,38] and it is

unchanged with respect to φ, i.e.∂Hr

∂φ= 0,

∂Hφ

∂φ= 0,

∂Hz

∂φ= 0[24]. Furthermore,

from electromagnetic theory, we know the magnetic field has only the azimuthal

component, so Hr, Hz are zero, i.e. �H = Hφ(r, z)eφ[19]. Simplifying equation

(3.5) yields

− ∂

∂z

(1

σ

∂H

∂z

)− ∂

∂r

(1

σr

∂(rH)

∂r

)= 0,

or∂

∂z

(1

σ

∂H

∂z

)+

∂

∂r

(1

σr

∂(rH)

∂r

)= 0,

or1

σ

∂2H

∂z2+

∂H

∂z

∂

∂z

(1

σ

)+

1

σ

[1

r

∂2(rH)

∂r2+

∂

∂r

(1

r

)∂(rH)

∂r

]= 0. (3.6)

In our study, we denote σ as a function of depth z only, i.e. σ = σ(z), and

we now have

∂2H

∂z2+ σ

∂H

∂z

∂

∂z

(1

σ

)+

1

r

∂2(rH)

∂r2+

∂

∂r

(1

r

)∂(rH)

∂r= 0,

or∂2H

∂z2+ σ

∂H

∂z

∂

∂z

(1

σ

)+

1

r

∂

∂r

(r∂H

∂r+H

)− 1

r2

(r∂H

∂r+H

)= 0,

or

∂2H

∂z2+ σ

∂H

∂z

∂

∂z

(1

σ

)+

1

r

(r∂2H

∂r2+

∂H

∂r+

∂H

∂r

)− 1

r

∂H

∂r− 1

r2H = 0,

or∂2H

∂z2+ σ

∂H

∂z

∂

∂z

(1

σ

)+

∂2H

∂r2+

1

r

∂H

∂r− 1

r2H = 0. (3.7)

The next step, we use finite element method to establish a numerical solu-

tion of our problem. We apply the Galerkin’s Method of Weighted Residuals to

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15

equation (3.7). Recall the Laplace equation in three-dimension with the electric

charge on the cylinder

∇2H = ΔH =1

r

∂H

∂r+

∂2H

∂r2+

1

r2∂2H

∂φ2+

∂2H

∂z2. (3.8)

Since the problem is axisymmetric, we have that H is independent of φ. Substi-

tuting equation (3.8) into (3.7), our problem becomes

ΔH + σ∂H

∂z

∂

∂z

(1

σ

)− 1

r2H = 0, (3.9)

where r ∈ [10, 190], z ∈ [0, 180]. The boundary condition (BC) of problem (3.9)

at different depths and distances from the source is shown in Figure 3.2.

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16

Figure 3.2: Boundary conditions of the Earth structure.

The values of magnetic field obtained by the receiver on the surface at

z = 0 m are from mathematical model of Sripunya, (2005) when b = 0.05 m−1.

The values of magnetic field decreases to zero as the depth increases, i.e we assume

that the values of magnetic field is zero at z = 180 m. The values of the magnetic

field on ∂Ω2 and ∂Ω4 are obtained from the approximation by linear functions,

H(r, z) = −8.83× 10−5z + 0.0159 T for ∂Ω4 and H(r, z) = −4.4× 10−5z + 0.0008

T for ∂Ω2, respectively.

After that we transform equation (3.9) into weak formulation to find H ∈H1. Let

V = { v ∈ H1 : v is a continuous function on Ω, ∂v∂r

and ∂v∂z

are piecewise continuous

on Ω and v = 0 on ∂Ω }.The weak formulation of equation (3.9)

(ΔH, v) +

(σ∂H

∂z

∂

∂z

( 1

σ

), v

)− (

1

r2H, v) = 0, v ∈ V,

or ∫Ω

ΔHvdΩ +

∫Ω

σ∂H

∂z

∂

∂z

( 1

σ

)vdΩ−

∫Ω

1

r2HvdΩ = 0. (3.10)

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17

By Green’s identity[32],∫Ω

ΔHvdΩ =

∫∂Ω

v∇H · �nds−∫Ω

∇H · ∇vdΩ

=

∫∂Ω1

v∇H · �nds+∫∂Ω2

v∇H · �nds+∫∂Ω3

v∇H · �nds

+

∫∂Ω4

v∇H · �nds−∫Ω

∇H · ∇vdΩ.

Since v ∈ V , v = 0 on ∂Ω1, ∂Ω2, ∂Ω3 and ∂Ω4. We have∫Ω

ΔHvdΩ = −∫Ω

∇H · ∇vdΩ. (3.11)

Substitute equation (3.11) into (3.10), we obtain

−∫Ω

∇H · ∇vdΩ +

∫Ω

σ∂H

∂z

∂

∂z

( 1

σ

)vdΩ−

∫Ω

1

r2HvdΩ = 0. (3.12)

Using cylindrical co-ordinates (r, φ, z) [21], problem (3.12) becomes

−∫Ω

r∇H ·∇vdrdφdz+

∫Ω

rσ∂H

∂z

∂

∂z

( 1

σ

)vdrdφdz−

∫Ω

r1

r2Hvdrdφdz = 0. (3.13)

Since the problem is axisymmetric and H has only the azimuthal component in

cylindrical coordinate, we divide (3.13) by 2π and derive the following formulation

in terms of cylindrical co-ordinates (r, z):

−∫Ω

r∇H · ∇vdrdz +

∫Ω

rσ∂H

∂z

∂

∂z

( 1

σ

)vdrdz −

∫Ω

1

rHvdrdz = 0, (3.14)

where Ω is the 2D cross-section of domain Ω (φ is fixed), i.e. Ω = {(r, z), 10 ≤r ≤ 190, 0 ≤ z ≤ 180}.

Next we consider the two dimensional domain of equation (3.14). By di-

viding the domain into rectangular elements, we discretize r into 9 subintervals

equally, discretize z into 9 subintervals equally and (ri, zj) is a node of Ω on the

non overlapping rectangles such that the horizontal and vertical edges of these

rectangles are parallel to the r and z coordinate axes,respectively, i.e.

ri = 10 + 20i , i = 0, . . . , 9,

zj = 20j , j = 0, . . . , 9.

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18

Since the form of equation (3.14) suggests that the finite elements can have

an arbitrary shape and position in space computing integrals over their element

domains is a bit tricky. To overcome this difficulty, one uses a projection method

which maps the coordinates of a well known reference element to the coordinates

of an arbitrary element in space. Computing an integral on the local reference

element (e.g. its area) is easy. One just has to capture the effect of the map-

ping (deformation, stretching, shearing) to get the right value of the integral for

the global element domain. During the mapping process the points in the local

coordinate system ξ, η (here: parent domain) get mapped to points in the global

coordinate system r, z by a mapping the values range from -1 to +1, and the ref-

erence coordinates are as (ξ1, η1) = (−1,−1), (ξ2, η2) = (1,−1) , (ξ3, η3) = (1, 1) ,

(ξ4, η4) = (−1, 1) that represent in Figure 3.3(b).

(a) A rectangular elements.

(b) The

reference

element.

Figure 3.3: The coordinate transformation (r, z) in terms of the local coordinates

(ξ, η)

For convenience, we rename the node (ri, zj) to Hi so we have {Hi}100i=1 as

the nodes of the element in Ω as shown in Figure 3.4.

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Figure 3.4: The nodes {Hi}100i=1 of the elements.

For simplicity and to avoid any confusion, we use H(Xi), i = 1, 2, . . . , 100

for Hi, i = 1, 2, . . . , 100. In other words, we define nodes Xi, i = 1, 2, . . . , 100 for

(ri, zj), i, j = 0, 1, . . . , 9.

For each i = 1, 2, . . . , 100 define ϕj as basis function such that

ϕj(Xi) =

{1 , i = j

0 , i �= j.

Figure 3.5: The basis function ϕj.

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The support of ϕj (the set of points X for which ϕj(X) �= 0) contains the

rectangles with the common node Xj (the shaded area in Figure 3.5). A function

v ∈ V can be written in the form of linear combination of basis function ϕi

v(X) =100∑i=1

αiϕi(X).

We obtain v(Xj) = αj by choosing appropriate values for αj. Equation (3.14)

becomes

−(r∇H,∇ϕi) +

(rσ

∂H

∂z

∂

∂z

( 1

σ

), ϕi

)− (

1

rH, ϕi) = 0.

Next, we consider the solution in the form of linear combination of basis

function ϕj

H(X) =100∑j=1

Hjϕj(X),

when Hj is the unknown parameters.

Then equation (3.14) can be written in the form of linear combination as

follows

100∑j=1

Hj

[−

∫Ω

r(∇ϕj · ∇ϕi)drdz +

∫Ω

r

(σ∂

∂z

( 1

σ

))∂ϕj

∂zϕidrdz

−∫Ω

1

rϕjϕidrdz

]= 0, (3.15)

for each i = 1, 2, . . . , 100.

Let Mi,j = −∫Ω


∫Ω

r

(σ∂

∂z

( 1

σ

))∂ϕj

∂zϕidrdz

−∫Ω

1

rϕjϕidrdz.

Then, equation (3.15) becomes

100∑j=1

HjMi,j = 0,

for each i = 1, 2, . . . , 100, that is,

H1M1,1 +H2M1,2 +H3M1,3 + . . .+H100M1,100 = 0

H1M2,1 +H2M2,2 +H3M2,3 + . . .+H100M2,100 = 0

......

H1M100,1 +H2M100,2 +H3M100,3 + . . .+H100M100,100 = 0.

สำนกหอ


21

We have a linear system in form of matrix as follows

⎡⎢⎢⎢⎢⎢⎣

M1,1 M1,2 . . . M1,100

M2,1 M2,2 . . . M2,100

......

......

M100,1 M100,2 . . . M100,100

⎤⎥⎥⎥⎥⎥⎦

⎡⎢⎢⎢⎢⎢⎣

H1

H2

...

H100

⎤⎥⎥⎥⎥⎥⎦ =

⎡⎢⎢⎢⎢⎢⎣0

0...

0

⎤⎥⎥⎥⎥⎥⎦

or MH = 0.

To find M , we need a transformation from an original element to a ref-

erence element as shown in Figure 3.3

r = rk +h

2(1 + ξ), dr =

h

2dξ,

z = zk +h

2(1 + η), dz =

h

2dη, (3.16)

where k = 0, 1, . . . , 8.

From Figure 3.3(b), the relationship between coordinate (r, z) to the basis

functions in coordinate (ξ, η) Ni, i = 1, 2, 3, 4 are

r = N1r1 +N2r2 +N3r3 +N4r4,

z = N1z1 +N2z2 +N3z3 +N4z4. (3.17)

The basis functions can be written in the form of ξ and η as follows

N1(ξ, η) =1

4(1− ξ)(1− η),

N2(ξ, η) =1

4(1 + ξ)(1− η),

N3(ξ, η) =1

4(1 + ξ)(1 + η),

N4(ξ, η) =1

4(1− ξ)(1 + η). (3.18)

สำนกหอ


Chapter 4

Numerical Results for the Case of

σ(z) = σ0e−bz

In this chapter, we consider numerical solution of problem having exponen-

tial conductivity σ(z) = σ0e−bz, where σ0 is positive constant and b may be either

positive or negative.

Figure 4.1: Geometric model of the Earth structure in the case of σ(z) = σ0e−bz.

Substituting σ(z) = σ0e−bz into equation (3.15), we obtain

100∑j=1

Hj

[−

∫Ω


∫Ω

rb∂ϕj

∂zϕidrdz −

∫Ω

1

rϕjϕidrdz

]= 0, (4.1)

for each i = 1, 2, . . . , 100 and

Mi,j = −∫Ω


∫Ω

rb∂ϕj

∂zϕidrdz −

∫Ω

1

rϕjϕidrdz.

22

สำนกหอ


23

Next, we consider the value of Mi,j. We separate Mi,j into 3 terms, i.e.

Mi,j = Ai,j + Bi,j + Ci,j where

Ai,j = −∫Ω

r∇ϕj · ∇ϕidrdz, Bi,j =

∫Ω

br∂ϕj

∂zϕidrdz, Ci,j = −

∫Ω

1

rϕjϕidrdz.

To calculate elements in Ai,j, Bi,j and Ci,j, we transform r, z to ξ, η, respec-

tively, by using the transformation equation (3.16) together with basis functions

equation (3.18).

Consider the value of −Ai,j =

∫Ω

r∇ϕj · ∇ϕidrdz by using Chain rule

and Jacobian transform thus

∫Ω

r∇ϕj · ∇ϕidrdz =

∫∫Ω

r

[∂ϕj

∂r

∂ϕi

∂r+

∂ϕj

∂z

∂ϕi

∂z

]drdz

=

∫ 1

−1

∫ 1

−1

r

[( ˆ∂ϕj

∂ξ

∂ξ

∂r+

ˆ∂ϕj

∂η

∂η

∂r

)( ˆ∂ϕi

∂ξ

∂ξ

∂r+

ˆ∂ϕi

∂η

∂η

∂r

)

+

( ˆ∂ϕj

∂ξ

∂ξ

∂z+

ˆ∂ϕj

∂η

∂η

∂z

)( ˆ∂ϕi

∂ξ

∂ξ

∂z+

ˆ∂ϕi

∂η

∂η

∂z

)](h

2

)2

dξdη

=

∫ 1

−1

∫ 1

−1

(2

h

)2

r

[ ˆ∂ϕj

∂ξ

ˆ∂ϕi

∂ξ+

ˆ∂ϕj

∂η

ˆ∂ϕi

∂η

](h

2

)2

dξdη

=

∫ 1

−1

∫ 1

−1

(rk +

h

2(1 + ξ)

)[ ˆ∂ϕj

∂ξ

ˆ∂ϕi

∂ξ+

ˆ∂ϕj

∂η

ˆ∂ϕi

∂η

]dξdη,

where k = 0, 1, . . . , 8 and ϕi, ϕj are basis functions in coordinate ξ, η for each

i, j = 1, 2, . . . , 100.

For the corresponding linear rectangular elements, the approximation∫ 1

−1

∫ 1

−1

(rk +

h

2(1+ ξ)

)[ ˆ∂ϕj

∂ξ

ˆ∂ϕi

∂ξ+

ˆ∂ϕj

∂η

ˆ∂ϕi

∂η

]dξdη can be divided into nine cases

(Figure. 3.3(b)).

สำนกหอ


24Tab

le4.1:

Thevalueof

∫ 1 −1

∫ 1 −1

( r k+

h 2(1

+ξ))[

ˆ∂ϕj

∂ξ

ˆ∂ϕi

∂ξ

+ˆ

∂ϕj

∂η

ˆ∂ϕi

∂η

] dξdη

incoordinates

(ξ,η)

Elemen

tsCases

∫ 1 −1

∫ 1 −1

( r k+

h 2(1

+ξ))[

ˆ∂ϕj

∂ξ

ˆ∂ϕi

∂ξ

+ˆ

∂ϕj

∂η

ˆ∂ϕi

∂η

] dξdη

inco

ord

inates(ξ,η

)Results

Solutions

∫ 1 −1

∫ 1 −1

( r k+

h 2(1

+ξ))[(

∂N

3

∂ξ

∂N

3

∂ξ

+∂N

3

∂η

∂N

3

∂η

) +

( ∂N

2

∂ξ

∂N

2

∂ξ

+∂N

2

∂η

∂N

2

∂η

)] dξdη

2

( (20+40k)

3+

5h

12

)

i=j

+

∫ 1 −1

∫ 1 −1

( r k+1+

h 2(1

+ξ))[(

∂N

4

∂ξ

∂N

4

∂ξ

+∂N

4

∂η

∂N

4

∂η

) +

( ∂N

1

∂ξ

∂N

1

∂ξ

+∂N

1

∂η

∂N

1

∂η

)] dξdη

+2

( (20+40(k

+1))

3+

h 4

)1 6(320(k

+1)+

8h)

i=j+

1

∫ 1 −1

∫ 1 −1

( r k+

h 2(1

+ξ))[(

∂N

4

∂ξ

∂N

3

∂ξ

+∂N

4

∂η

∂N

3

∂η

) +

( ∂N

1

∂ξ

∂N

2

∂ξ

+∂N

1

∂η

∂N

2

∂η

)] dξdη

−2( (5

+10k)

3+

h 12

)−

1 6

( (20+

20k)+

h

)

i=j-1

∫ 1 −1

∫ 1 −1

( r k+

h 2(1

+ξ))[(

∂N

3

∂ξ

∂N

4

∂ξ

+∂N

3

∂η

∂N

4

∂η

) +

( ∂N

2

∂ξ

∂N

1

∂ξ

+∂N

2

∂η

∂N

1

∂η

)] dξdη

−2( (5

+10k)

3+

h 12

)−

1 6

( (20+

20k)+

h

)

∫ 1 −1

∫ 1 −1

( r k+

h 2(1

+ξ))[(

∂N

3

∂ξ

∂N

2

∂ξ

+∂N

3

∂η

∂N

2

∂η

)] dξdη

i=j+

M+

∫ 1 −1

∫ 1 −1

( r k+1+

h 2(1

+ξ))[(

∂N

4

∂ξ

∂N

1

∂ξ

+∂N

4

∂η

∂N

1

∂η

)] dξdη

−( (5+10k)

3+

h 6

) −(5

+10(k

+1))

3−

1 6(40(k

+1)+

h)

∫ 1 −1

∫ 1 −1

( r k+

h 2(1

+ξ))[(

∂N

2

∂ξ

∂N

3

∂ξ

+∂N

2

∂η

∂N

3

∂η

)] dξdη

i=j-M

+

∫ 1 −1

∫ 1 −1

( r k+1+

h 2(1

+ξ))[(

∂N

1

∂ξ

∂N

4

∂ξ

+∂N

1

∂η

∂N

4

∂η

)] dξdη

−( (5+10k)

3+

h 6

) −(5

+10(k

+1))

3−

1 6(40(k

+1)+

h)

i=j+

M+1

∫ 1 −1

∫ 1 −1

( r k+

h 2(1

+ξ))[(

∂N

4

∂ξ

∂N

2

∂ξ

+∂N

4

∂η

∂N

2

∂η

)] dξdη

−1 6

( (20+

40k)+

h

)−

1 6

( (20+

40k)+

h

)

i=j-M-1

∫ 1 −1

∫ 1 −1

( r k+

h 2(1

+ξ))[(

∂N

2

∂ξ

∂N

4

∂ξ

+∂N

2

∂η

∂N

4

∂η

)] dξdη

−1 6

( (20+

40k)+

h

)−

1 6

( (20+

40k)+

h

)

i=j+

M-1

∫ 1 −1

∫ 1 −1

( r k+

h 2(1

+ξ))[(

∂N

3

∂ξ

∂N

1

∂ξ

+∂N

3

∂η

∂N

1

∂η

)] dξdη

−1 6

( (20+

40k)+

h

)−

1 6

( (20+

40k)+

h

)

i=j-M+1

∫ 1 −1

∫ 1 −1

( r k+

h 2(1

+ξ))[(

∂N

1

∂ξ

∂N

3

∂ξ

+∂N

1

∂η

∂N

3

∂η

)] dξdη

−1 6

( (20+

40k)+

h

)−

1 6

( (20+

40k)+

h

)

สำนกหอ


25

where k = 0, 1, . . . , 8 for each i, j = 1, 2, . . . , 100.

Then, from the value of

∫ 1

−1

∫ 1

−1

(rk+

h

2(1+ ξ)

)[ ˆ∂ϕj

∂ξ

ˆ∂ϕi

∂ξ+

ˆ∂ϕj

∂η

ˆ∂ϕi

∂η

]dξdη can

be written in the form of matrix as follows

−

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

A6 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0A3 A1 s1 A2 s2 0 0 0 0 0 0 0 0 0 0 0 00 0 t 0 0 0 0 0 0 0 0 0 0 0 0 0 0A4 A2 s1 A1 s1 A2 s3 0 0 0 0 0 0 0 0 0 00 0 0 0 t 0 0 0 0 0 0 0 0 0 0 0 00 0 s4 A2 s1 A1 s1 A2 s3 0 0 0 0 0 0 0 00 0 0 0 0 0 t 0 0 0 0 0 0 0 0 0 00 0 0 0 s4 A2 s1 A1 s1 A2 s3 0 0 0 0 0 00 0 0 0 0 0 0 0 t 0 0 0 0 0 0 0 00 0 0 0 0 0 s4 A2 s1 A1 s1 A2 s3 0 0 0 00 0 0 0 0 0 0 0 0 0 t 0 0 0 0 0 00 0 0 0 0 0 0 0 s4 A2 s1 A1 s1 A2 s3 0 00 0 0 0 0 0 0 0 0 0 0 0 t 0 0 0 00 0 0 0 0 0 0 0 0 0 s4 A2 s1 A1 s1 A2 A50 0 0 0 0 0 0 0 0 0 0 0 0 0 t 0 00 0 0 0 0 0 0 0 0 0 0 0 s4 A2 s1 A1 A50 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 A6

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

ub0u1ub1u2ub2u3ub3u4ub4u5ub5u6ub6u7ub7u8ub8

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

= 0

or AU = 0,

where

A = −

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

A6 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0A3 A1 s1 A2 s2 0 0 0 0 0 0 0 0 0 0 0 00 0 t 0 0 0 0 0 0 0 0 0 0 0 0 0 0A4 A2 s1 A1 s1 A2 s3 0 0 0 0 0 0 0 0 0 00 0 0 0 t 0 0 0 0 0 0 0 0 0 0 0 00 0 s4 A2 s1 A1 s1 A2 s3 0 0 0 0 0 0 0 00 0 0 0 0 0 t 0 0 0 0 0 0 0 0 0 00 0 0 0 s4 A2 s1 A1 s1 A2 s3 0 0 0 0 0 00 0 0 0 0 0 0 0 t 0 0 0 0 0 0 0 00 0 0 0 0 0 s4 A2 s1 A1 s1 A2 s3 0 0 0 00 0 0 0 0 0 0 0 0 0 t 0 0 0 0 0 00 0 0 0 0 0 0 0 s4 A2 s1 A1 s1 A2 s3 0 00 0 0 0 0 0 0 0 0 0 0 0 t 0 0 0 00 0 0 0 0 0 0 0 0 0 s4 A2 s1 A1 s1 A2 A50 0 0 0 0 0 0 0 0 0 0 0 0 0 t 0 00 0 0 0 0 0 0 0 0 0 0 0 s4 A2 s1 A1 A50 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 A6

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

and U = [ ub0 u1 ub1 u2 ub2 u3 ub3 u4 ub4 u5 ub5 u6 ub6 u7 ub7 u8 ub8 ]T

such that ub0 = [H1 H2 H3 H4 H5 H6 H7 H8 H9 H10 H11 ]T ,

ub8 = [H90 H91 H92 H93 H94 H95 H96 H97 H98 H99 H100 ]T ,

ui =[H10i+2 H10i+3 H10i+4 H10i+5 H10i+6 H10i+7 H10i+8 H10i+9

]T,

and ubj−1 =[H10j H10j+1

]T, for each i = 1, 2, . . . , 8 and j = 2, 3, . . . , 8 and

สำนกหอ


26

A1=

1 6

⎡ ⎢ ⎢ ⎢ ⎢ ⎣320+8h

−60−

h0

00

00

0−6

0−h

640+8h

−100

−h0

00

00

0−1

00−h

960+8h

−140−h

00

00

00

−140

−h1280+8h

−180

−h0

00

00

0−1

80−h

1600+8h

−220

−h0

00

00

0−2

20−

h1920+8h

−260

−h0

00

00

0−2

60−h

2240+8h

−300

−h0

00

00

0−3

00−h

2560+8h

⎤ ⎥ ⎥ ⎥ ⎥ ⎦,

A2=

1 6

⎡ ⎢ ⎢ ⎢ ⎢ ⎣−40−h

−60−

h0

00

00

0−6

0−h

−80−h

−100

−h0

00

00

0−1

00−h

−120

−h−1

40−h

00

00

00

−140

−h−1

60−

h−1

80−h

00

00

00

−180

−h−2

00−

h−2

20−h

00

00

00

−220−

h−2

40−h

−260

−h0

00

00

0−2

60−h

−280

−h−3

00−h

00

00

00

−300

−h−3

20−h

⎤ ⎥ ⎥ ⎥ ⎥ ⎦,

A3=

1 6

⎡ ⎢ ⎢ ⎢ ⎢ ⎣(−20−h

)(−

40−h

)(−

60−h

)0

00

00

00

(−20−h

)0

(−60−h

)(−

80−h

)(−

100−h

)0

00

00

00

00

(−100−h

)(−

120−h

)(−

140−h

)0

00

00

00

00

(−140−h

)(−

160−h

)(−

180−h

)0

00

00

00

00

(−180−h

)(−

200−h

)(−

220−h

)0

00

00

00

00

(−220−h

)(−

240−h

)(−

260−h

)0

00

00

00

00

(−260−h

)(−

280−h

)(−

300−h

)0

00

00

00

00

(−300−h

)(−

320−h

)(−

340−h

)0

⎤ ⎥ ⎥ ⎥ ⎥ ⎦,

A4=

1 6

⎡ ⎢ ⎢ ⎢ ⎣0000000000(−

20−h

)0000000000

00000000000

00000000000

00000000000

00000000000

00000000000

00000000000

0

⎤ ⎥ ⎥ ⎥ ⎦,A

5=

1 6

⎡ ⎢ ⎢ ⎢ ⎣0

0000000000

00000000000

00000000000

00000000000

00000000000

00000000000

00000000000

(−34

0−h)0000000000

⎤ ⎥ ⎥ ⎥ ⎦,A

6=

⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣10000000000

01000000000

00100000000

00010000000

00001000000

00000100000

00000010000

00000001000

00000000100

00000000010

00000000001

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦,

สำนกหอ


27

s1 =16

⎡⎢⎢⎣

0 (−20−h)0 00 00 00 00 00 0

(−340−h) 0

⎤⎥⎥⎦ , s2 =

16

⎡⎢⎢⎣

(−340−h) 00 00 00 00 00 00 00 0

⎤⎥⎥⎦ , s3 =

16

⎡⎢⎢⎣

0 00 00 00 00 00 00 0

(−340−h) 0

⎤⎥⎥⎦,

s4 =16

⎡⎢⎢⎣

0 (−20−h)0 00 00 00 00 00 00 0

⎤⎥⎥⎦ , t = [ 1 0

0 1 ].

Since we know the value of the magnetic field at the boundaryH1, H2, H3, . . . , H11,

H90, H91, H92, . . . , H100 and H10j, H10j+1 for all j = 2, 3, . . . , 8. The system can be

written as

−

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

A1 A2 0 0 0 0 0 0

A2 A1 A2 0 0 0 0 0

0 A2 A1 A2 0 0 0 0

0 0 A2 A1 A2 0 0 0

0 0 0 A2 A1 A2 0 0

0 0 0 0 A2 A1 A2 0

0 0 0 0 0 A2 A1 A2

0 0 0 0 0 0 A2 A1

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

u1

u2

u3

u4

u5

u6

u7

u8

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

=

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

a1

a2

a3

a4

a5

a6

a7

a8

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

or AU = P ,

where

A = −

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

A1 A2 0 0 0 0 0 0

A2 A1 A2 0 0 0 0 0

0 A2 A1 A2 0 0 0 0

0 0 A2 A1 A2 0 0 0

0 0 0 A2 A1 A2 0 0

0 0 0 0 A2 A1 A2 0

0 0 0 0 0 A2 A1 A2

0 0 0 0 0 0 A2 A1

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

and U = ui =[H10i+2 H10i+3 H10i+4 H10i+5 H10i+6 H10i+7 H10i+8 H10i+9

]T, P =

[a1, . . . , a8

]Tfor all i = 1, 2, . . . , 8 and

สำนกหอ


28

a1 =16

⎡⎢⎢⎢⎣

(−20−h)(0.0159)+(−40−h)(0.0053)+(−60−h)(0.0032)+(−20−h)(0.0141)+(−20−h)(0.0124)(−60−h)(0.0053)+(−80−h)(0.0032)+(−100−h)(0.0023)

(−100−h)(0.0032)+(−120−h)(0.0023)+(−140−h)(0.0018)(−140−h)(0.0023)+(−160−h)(0.0018)+(−180−h)(0.0014)(−180−h)(0.0018)+(−200−h)(0.0014)+(−220−h)(0.0012)(−220−h)(0.0014)+(−240−h)(0.0012)+(−260−h)(0.0011)(−260−h)(0.0012)+(−280−h)(0.0011)+(−300−h)(0.0009)

(−300−h)(0.0011)+(−320−h)(0.0009)+(−340−h)(0.0008)+(−340−h)(0.0007)+(−340−h)(0.0006)

⎤⎥⎥⎥⎦,

a2 =16

⎡⎢⎢⎣

(−20−h)(0.0141)+(−20−h)(0.0124)+(−20−h)(0.0106)000000

(−340−h)(0.0007)+(−340−h)(0.0006)+(−340−h)(0.0005)

⎤⎥⎥⎦,

a3 =16

⎡⎢⎢⎣

(−20−h)(0.0124)+(−20−h)(0.0106)+(−20−h)(0.0088)000000

(−340−h)(0.0006)+(−340−h)(0.0005)+(−340−h)(0.00045)

⎤⎥⎥⎦,

a4 =16

⎡⎢⎢⎣

(−20−h)(0.0106)+(−20−h)(0.0088)+(−20−h)(0.0071)000000

(−340−h)(0.0005)+(−340−h)(0.00045)+(−340−h)(0.0004)

⎤⎥⎥⎦,

a5 =16

⎡⎢⎢⎣

(−20−h)(0.0088)+(−20−h)(0.0071)+(−20−h)(0.0053)000000

(−340−h)(0.00045)+(−340−h)(0.0004)+(−340−h)(0.0003)

⎤⎥⎥⎦,

a6 =16

⎡⎢⎢⎣

(−20−h)(0.0071)+(−20−h)(0.0053)+(−20−h)(0.0035)000000

(−340−h)(0.0004)+(−340−h)(0.0003)+(−340−h)(0.0002)

⎤⎥⎥⎦,

a7 =16

⎡⎢⎢⎣

(−20−h)(0.0053)+(−20−h)(0.0035)+(−20−h)(0.0018)000000

(−340−h)(0.0003)+(−340−h)(0.0002)+(−340−h)(0.0001)

⎤⎥⎥⎦,

a8 =16

⎡⎢⎢⎣

(−20−h)(0.0035)+(−20−h)(0.0018)000000

(−340−h)(0.0002)+(−340−h)(0.0001)

⎤⎥⎥⎦.

สำนกหอ


29

Next, consider the value of Bi,j =

∫Ω

br∂ϕj

∂zϕidrdz by using Chain rule

and Jacobian transform thus∫Ω

br∂ϕj

∂zϕidrdz =

∫ 1

−1

∫ 1

−1

br

[ ˆ∂ϕj

∂ξ

∂ξ

∂z+

ˆ∂ϕj

∂η

∂η

∂z

]ϕi

(h2

)2

dξdη

=

∫ 1

−1

∫ 1

−1

br

[ ˆ∂ϕj

∂ξ0 +

ˆ∂ϕj

∂η

2

h

]ϕi

(h2

)2

dξdη

=bh

2

∫ 1

−1

∫ 1

−1

rˆ∂ϕj

∂ηϕidξdη

=bh

2

∫ 1

−1

∫ 1

−1

(rk +

h

2(1 + ξ)

) ˆ∂ϕj

∂ηϕidξdη,

where k = 0, 1, . . . , 8 and ϕi, ϕj are basis functions in coordinate ξ, η for each

i, j = 1, 2, . . . , 100.

For the corresponding linear rectangular elements, the approximationbh

2

∫ 1

−1

∫ 1

−1

(rk +

h

2(1 + ξ)

) ˆ∂ϕj

∂ηϕidξdη can be divided into nine cases (Figure.

3.3(b)).

สำนกหอ


30Table 4.2: The value of

bh

2

∫ 1

−1

∫ 1

−1

(rk +

h

2(1 + ξ)

) ˆ∂ϕj

∂ηϕidξdη in coordinates (ξ, η)

Elements Casesbh

2

∫ 1

−1

∫ 1

−1

(rk +

h

2(1 + ξ)

) ˆ∂ϕj

∂ηϕidξdη in coordinates (ξ, η) Results Solutions

bh

2

∫ 1

−1

∫ 1

−1

[(rk +

h

2(1 + ξ)

)(∂N3

∂ηN3 +

∂N2

∂ηN2

)]dξdη bh

32

[(160(2k+1)

3+ 4h

)−

(160(2k+1)

3+ 4h

)]

i=j +bh

2

∫ 1

−1

∫ 1

−1

[(rk+1 +

h

2(1 + ξ)

)(∂N4

∂ηN4 +

∂N1

∂ηN1

)]dξdη + bh

32

[(160(2k+3)

3+ 4h

3

)−

(160(2k+3)

3+ 4h

3

)]0

i=j+1bh

2

∫ 1

−1

∫ 1

−1

(rk +

h

2(1 + ξ)

)[(∂N4

∂ηN3 +

∂N1

∂ηN2

)]dξdη bh

32

[((80+160k)

3+ 4h

3

)−

((80+160k)

3+ 4h

3

)]0

i=j-1bh

2

∫ 1

−1

∫ 1

−1

(rk +

h

2(1 + ξ)

)[(∂N3

∂ηN4 +

∂N2

∂ηN1

)]dξdη bh

32

[((80+160k)

3+ 4h

3

)−

((80+160k)

3+ 4h

3

)]0

bh

2

∫ 1

−1

∫ 1

−1

[(rk +

h

2(1 + ξ)

)(∂N3

∂ηN2

)]dξdη

i=j+M +bh

2

∫ 1

−1

∫ 1

−1

[(rk+1 +

h

2(1 + ξ)

)(∂N4

∂ηN1

)]dξdη bh

32

[(160(2k+1)

3+ 4h

)+

(160(2k+3)

3+ 4h

3

)]bh96

(640(k + 1) + 16h)

bh

2

∫ 1

−1

∫ 1

−1

[(rk +

h

2(1 + ξ)

)(∂N2

∂ηN3

)]dξdη

i=j-M +bh

2

∫ 1

−1

∫ 1

−1

[(rk+1 +

h

2(1 + ξ)

)(∂N1

∂ηN4

)]dξdη bh

32

[−

(160(2k+1)

3+ 4h

)−

(160(2k+3)

3+ 4h

3

)]− bh

96(640(k + 1) + 16h)

i=j+M+1bh

2

∫ 1

−1

∫ 1

−1

(rk +

h

2(1 + ξ)

)(∂N4

∂ηN2

)dξdη bh

32

(80(2k+1)

3+ 4h

3

)bh96

(80(2k + 1) + 4h)

i=j-M-1bh

2

∫ 1

−1

∫ 1

−1

(rk +

h

2(1 + ξ)

)(∂N2

∂ηN4

)dξdη − bh

32

(80(2k+1)

3+ 4h

3

)− bh

96(80(2k + 1) + 4h)

i=j+M-1bh

2

∫ 1

−1

∫ 1

−1

(rk +

h

2(1 + ξ)

)(∂N3

∂ηN1

)dξdη bh

32

(80(2k+1)

3+ 4h

3

)bh96

(80(2k + 1) + 4h)

i=j-M+1bh

2

∫ 1

−1

∫ 1

−1

(rk +

h

2(1 + ξ)

)(∂N1

∂ηN3

)dξdη − bh

32

(80(2k+1)

3+ 4h

3

)− bh

96(80(2k + 1) + 4h)

สำนกหอ


31

where k = 0, 1, . . . , 8 for each i, j = 1, 2, . . . , 100.

Then, from the value ofbh

2

∫ 1

−1

∫ 1

−1

(rk +

h

2(1 + ξ)

) ˆ∂ϕj

∂ηϕidξdη can be written

in the form of matrix as follows

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

B1 B2 0 0 0 0 0 0

B3 B1 B2 0 0 0 0 0

0 B3 B1 B2 0 0 0 0

0 0 B3 B1 B2 0 0 0

0 0 0 B3 B1 B2 0 0

0 0 0 0 B3 B1 B2 0

0 0 0 0 0 B3 B1 B2

0 0 0 0 0 0 B3 B1

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

u1

u2

u3

u4

u5

u6

u7

u8

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

= −

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

d1

d2

d3

d4

d5

d6

d7

d8

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

or BU = Q,

where

B =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

B1 B2 0 0 0 0 0 0

B3 B1 B2 0 0 0 0 0

0 B3 B1 B2 0 0 0 0

0 0 B3 B1 B2 0 0 0

0 0 0 B3 B1 B2 0 0

0 0 0 0 B3 B1 B2 0

0 0 0 0 0 B3 B1 B2

0 0 0 0 0 0 B3 B1

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦


]T, Q = −

[d1, . . . , d8

]Tfor all i = 1, 2, . . . , 8 and

B1 =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦,

สำนกหอ


32

B2 =bh96

⎡⎢⎢⎢⎣

640+16h 240+4h 0 0 0 0 0 0240+4h 1280+16h 400+4h 0 0 0 0 0

0 400+4h 1920+16h 560+4h 0 0 0 00 0 560+4h 2560+16h 720+4h 0 0 00 0 0 720+4h 3200+16h 880+4h 0 00 0 0 0 880+4h 3840+16h 1040+4h 00 0 0 0 0 1040+4h 4480+16h 1200+4h0 0 0 0 0 0 1200+4h 5120+16h

⎤⎥⎥⎥⎦,

B3 =bh96

⎡⎢⎢⎢⎢⎣

−(640+16h) −(240+4h) 0 0 0 0 0 0−(240+4h) −(1280+16h) −(400+4h) 0 0 0 0 0

0 −(400+4h) −(1920+16h) −(560+4h) 0 0 0 00 0 −(560+4h) −(2560+16h) −(720+4h) 0 0 00 0 0 −(720+4h) −(3200+16h) −(880+4h) 0 00 0 0 0 −(880+4h) −(3840+16h) −(1040+4h) 00 0 0 0 0 −(1040+4h) (4480+16h) −(1200+4h)

0 0 0 0 0 0 −(1200+4h) −(5120+16h)

⎤⎥⎥⎥⎥⎦,

d1 =bh96

⎡⎢⎢⎢⎢⎣

−(80+4h)(0.0159)−(640+16h)(0.0053)−(240+4h)(0.0032)+(80+4h)(0.0124)−(240+4h)(0.0053)−(1280+16h)(0.0032)−(400+4h)(0.0023)−(400+4h)(0.0032)−(1920+16h)(0.0023)−(560+4h)(0.0018)−(560+4h)(0.0023)−(2560+16h)(0.0018)−(720+4h)(0.0014)−(720+4h)(0.0018)−(3200+16h)(0.0014)−(880+4h)(0.0012)−(880+4h)(0.0014)−(3840+16h)(0.0012)−(1040+4h)(0.0011)−(1040+4h)(0.0012)−(4480+16h)(0.0011)−(1200+4h)(0.0009)

−(1200+4h)(0.0011)−(5120+16h)(0.0009)−(1360+4h)(0.0008)+(1360+4h)(0.0006)

⎤⎥⎥⎥⎥⎦,

d2 =bh96

⎡⎢⎢⎣

−(80+4h)(0.0141)+(80+4h)(0.0106)000000

−(1360+4h)(0.0007)+(1360+4h)(0.0005)

⎤⎥⎥⎦ , d3 =

bh96

⎡⎢⎢⎣

−(80+4h)(0.0124)+(80+4h)(0.0088)000000

−(1360+4h)(0.0006)+(1360+4h)(0.00045)

⎤⎥⎥⎦,

d4 =bh96

⎡⎢⎢⎣

−(80+4h)(0.0106)+(80+4h)(0.0071)000000

−(1360+4h)(0.0005)+(1360+4h)(0.0004)

⎤⎥⎥⎦ , d5 =

bh96

⎡⎢⎢⎣

−(80+4h)(0.0088)+(80+4h)(0.0053)000000

−(1360+4h)(0.00045)+(1360+4h)(0.0003)

⎤⎥⎥⎦,

สำนกหอ


33

d6 =bh96

⎡⎢⎣

−(80+4h)(0.0071)+(80+4h)(0.0035)000000

−(1360+4h)(0.0004)+(1360+4h)(0.0002)

⎤⎥⎦ , d7 =

bh96

⎡⎢⎣

−(80+4h)(0.0053)+(80+4h)(0.0018)000000

−(1360+4h)(0.0003)+(1360+4h)(0.0001)

⎤⎥⎦,

d8 =bh96

⎡⎢⎢⎣

−(80+4h)(0.0035)000000

−(1360+4h)(0.0002)

⎤⎥⎥⎦.

Next, consider the value of −Ci,j =

∫Ω

1

rϕjϕidrdz by using Jacobian

transform thus ∫Ω

1

rϕjϕidrdz =

h2

4

∫ 1

−1

∫ 1

−1

1

rϕjϕidξdη

=h2

4

∫ 1

−1

∫ 1

−1

ϕjϕi(rk +

h2(1 + ξ)

)dξdη,where k = 0, 1, . . . , 8 and ϕi, ϕj are basis functions in coordinate ξ, η for each

i, j = 1, 2, . . . , 100.

For the corresponding linear rectangular elements, the approximationh2

4

∫ 1

−1

∫ 1

−1

ϕjϕi(rk +

h2(1 + ξ)

)dξdη can be divided into nine cases (Figure. 3.3(b)).

สำนกหอ


34Table 4.3: The value of

h2

4

∫ 1

−1

∫ 1

−1

ϕjϕi(rk +

h2(1 + ξ)

)dξdη in coordinates (ξ, η)

Elements Casesh2

4

∫ 1

−1

∫ 1

−1

ϕj ϕi(rk + h

2(1 + ξ)

)dξdη in coordinates (ξ, η) Results Solutions

i=jh2

4

∫ 1

−1

∫ 1

−1

[(N3N3 +N2N2)(rk + h

2(1 + ξ)

) +(N4N4 +N1N1)(rk+1 + h

2(1 + ξ)

)]dξdη h2

64

(8

3h3 (4pk+1) +8

3h3 (4pk+1)

)1

12h(4pk+1)

i=j+1h2

4

∫ 1

−1

∫ 1

−1

(N4N3 +N1N2)(rk + h

2(1 + ξ)

)dξdη h2

64

(2(

83h3

)(4qk+1)

)1

12h(4qk+1)

i=j-1h2

4

∫ 1

−1

∫ 1

−1

(N3N4 +N2N1)(rk + h

2(1 + ξ)

)dξdη h2

64

(2(

83h3

)(4qk+1)

)1

12h(4qk+1)

i=j+Mh2

4

∫ 1

−1

∫ 1

−1

[N3N2(

rk + h2(1 + ξ)

) +N4N1(

rk+1 + h2(1 + ξ)

)]dξdη h2

64

(4

3h3 (4pk+1) +4

3h3 (4pk+1)

)1

12h(pk+1)

i=j-Mh2

4

∫ 1

−1

∫ 1

−1

[N2N3(

rk + h2(1 + ξ)

) +N1N4(

rk+1 + h2(1 + ξ)

)]dξdη h2

64

(4

3h3 (4pk+1) +4

3h3 (4pk+1)

)1

12h(pk+1)

i=j+M+1h2

4

∫ 1

−1

∫ 1

−1

N4N2(rk + h

2(1 + ξ)

)dξdη h2

64

(4

3h3 (4qk+1)

)1

12h(qk+1)

i=j-M-1h2

4

∫ 1

−1

∫ 1

−1

N2N4(rk + h

2(1 + ξ)

)dξdη h2

64

(4

3h3 (4qk+1)

)1

12h(qk+1)

i=j+M-1h2

4

∫ 1

−1

∫ 1

−1

N3N1(rk + h

2(1 + ξ)

)dξdη h2

64

(4

3h3 (4qk+1)

)1

12h(qk+1)

i=j-M+1h2

4

∫ 1

−1

∫ 1

−1

N1N3(rk + h

2(1 + ξ)

)dξdη h2

64

(4

3h3 (4qk+1)

)1

12h(qk+1)

สำนกหอ


35

where k = 0, 1, . . . , 8 for each i, j = 1, 2, . . . , 100 and

pk+1 =

[−

((20 + 40k)h− h2 + (200 + 1600x) ln(2k + 1) + (200 + 1600x) ln(2)

+(200 + 1600x) ln(5)− (200 + 1600x) ln(10(2k + 1) + h))]

+

[−

((60 + 40k)h+ 3h2 + ln(2k + 3)(2h2 + (120 + 80k)h+ (1800 + 1600y))

+ ln(2)(2h2 + (120 + 80k)h+ (1800 + 1600y))

+ ln(5)(2h2 + (120 + 80k)h+ (1800 + 1600y))

− ln(10(2k + 3) + h)(2h2 + (120 + 80k)h+ (1800 + 1600y)))]

,

qk+1 =

[((20 + 40k)h+ h2 + ln(2k + 1)((20 + 40k)h+ (200 + 1600x))

+ ln(2)((20 + 40k)h+ (200 + 1600x)) + ln(5)((20 + 40k)h+ (200 + 1600x))

− ln(10(2k + 1) + h)((20 + 40k)h+ (200 + 1600x)))]

,

for each x = 0, 1, 3, 6, 10, 15, 21, 28 and 36 and y = 0, 2, 5, 9, 14, 20, 27 and 35.

Then, from the value ofh2

4

∫ 1

−1

∫ 1

−1

ϕjϕi(rk +

h2(1 + ξ)

)dξdη can be written in the

form of matrix as follows

−

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

C1 C2 0 0 0 0 0 0

C2 C1 C2 0 0 0 0 0

0 C2 C1 C2 0 0 0 0

0 0 C2 C1 C2 0 0 0

0 0 0 C2 C1 C2 0 0

0 0 0 0 C2 C1 C2 0

0 0 0 0 0 C2 C1 C2

0 0 0 0 0 0 C2 C1

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

u1

u2

u3

u4

u5

u6

u7

u8

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

=

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

c1

c2

c3

c4

c5

c6

c7

c8

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

or CU = R,

where

สำนกหอ


36

C = −

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

C1 C2 0 0 0 0 0 0

C2 C1 C2 0 0 0 0 0

0 C2 C1 C2 0 0 0 0

0 0 C2 C1 C2 0 0 0

0 0 0 C2 C1 C2 0 0

0 0 0 0 C2 C1 C2 0

0 0 0 0 0 C2 C1 C2

0 0 0 0 0 0 C2 C1

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦


]T, R =

[c1, . . . , c8

]Tfor all i = 1, 2, . . . , 8 and

C1 =1

12h

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

4(p1) 4(q2) 0 0 0 0 0 0

4(q2) 4(p2) 4(q3) 0 0 0 0 0

0 4(q3) 4(p3) 4(q4) 0 0 0 0

0 0 4(q4) 4(p4) 4(q5) 0 0 0

0 0 0 4(q5) 4(p5) 4(q6) 0 0

0 0 0 0 4(q6) 4(p6) 4(q7) 0

0 0 0 0 0 4(q7) 4(p7) 4(q8)

0 0 0 0 0 0 4(q8) 4(p8)

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

,

C2 =1

12h

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

p1 q2 0 0 0 0 0 0

q2 p2 q3 0 0 0 0 0

0 q3 p3 q4 0 0 0 0

0 0 q4 p4 q5 0 0 0

0 0 0 q5 p5 q6 0 0

0 0 0 0 q6 p6 q7 0

0 0 0 0 0 q7 p7 q8

0 0 0 0 0 0 q8 p8

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

,

c1 =1

12h

⎡⎢⎢⎢⎢⎣

(q1)(0.0159)+(p1)(0.0053)+(q2)(0.0032)+4(q1)(0.0141)+(q1)(0.0124)(q2)(0.0053)+(p2)(0.0032)+(q3)(0.0023)(q3)(0.0032)+(p3)(0.0023)+(q4)(0.0018)(q4)(0.0023)+(p4)(0.0018)+(q5)(0.0014)(q5)(0.0018)+(p5)(0.0014)+(q6)(0.0012)(q6)(0.0014)+(p6)(0.0012)+(q7)(0.0011)(q7)(0.0012)+(p7)(0.0011)+(q8)(0.0009)

(q8)(0.0011)+(p8)(0.0009)+(q9)(0.0008)+4(q9)(0.0007)+(q9)(0.0006)

⎤⎥⎥⎥⎥⎦,

สำนกหอ


37

c2 =1

12h

⎡⎢⎣

(q1)(0.0141)+4(q1)(0.0124)+(q1)(0.0106)000000

(q9)(0.0007)+4(q9)(0.0006)+(q9)(0.0005)

⎤⎥⎦ , c3 =

112h

⎡⎢⎣

(q1)(0.0124)+4(q1)(0.0106)+(q1)(0.0088)000000

(q9)(0.0006)+4(q9)(0.0005)+(q9)(0.00045)

⎤⎥⎦,

c4 =1

12h

⎡⎢⎣

(q1)(0.0106)+4(q1)(0.0088)+(q1)(0.0071)000000

(q9)(0.0005)+4(q9)(0.00045)+(q9)(0.0004)

⎤⎥⎦ , c5 =

112h

⎡⎢⎣

(q1)(0.0088)+4(q1)(0.0071)+(q1)(0.0053)000000

(q9)(0.00045)+4(q9)(0.0004)+(q9)(0.0003)

⎤⎥⎦,

c6 =1

12h

⎡⎢⎣

(q1)(0.0071)+4(q1)(0.0053)+(q1)(0.0035)000000

(q9)(0.0004)+4(q9)(0.0003)+(q9)(0.0002)

⎤⎥⎦ , c7 =

112h

⎡⎢⎣

(q1)(0.0053)+4(q1)(0.0035)+(q1)(0.0018)000000

(q9)(0.0003)+4(q9)(0.0002)+(q9)(0.0001)

⎤⎥⎦,

c8 =1

12h

⎡⎢⎢⎣

(q1)(0.0035)+4(q1)(0.0018)000000

(q9)(0.0002)+4(q9)(0.0001)

⎤⎥⎥⎦.

Therefore equation (4.1) can be written in the form of matrix as follows

(−A+ B − C

)

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

u1

u2

u3

u4

u5

u6

u7

u8

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

=

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

a1 + d1 + c1

a2 + d2 + c2

a3 + d3 + c3

a4 + d4 + c4

a5 + d5 + c5

a6 + d6 + c6

a7 + d7 + c7

a8 + d8 + c8

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

or(−A+ B − C

)ui =

(a+ d+ c

),

where ui =[H10i+2 H10i+3 H10i+4 H10i+5 H10i+6 H10i+7 H10i+8 H10i+9

]T,

P =[a1, . . . , a8

]T, Q =

[d1, . . . , d8

]Tand R =

[c1, . . . , c8

]Tfor all i = 1, 2, . . . , 8.

4.1 Case of an Exponentially Decreasing Conductivity

Since the Galerkin’s Method of Weighted Residuals was applied to equation

สำนกหอ


38

(3.7), we obtained the values of magnetic field at various positions of the earth’s

structure with one layer having exponentially decreasing conductivity σ = σ0e−bz,

where b > 0. There is a source providing a DC voltage and a receiver on the

ground surface which picks up the signal from r = 10 m to r = 190 m. We

discrete the depth into 9 subintervals equally of the size h = 20 m, i.e. we consider

z = 0, 20, . . . , 180 m. We use constant b = 0.001, 0.01, 0.05, 0.075 and 0.1 m−1.

The numerical solutions of the magnetic field at each node is calculated by using

MATLAB program.

สำนกหอ


39

The values of magnetic field when b = 0.001 m−1 are computed as

H12 = 0.0051 H13 = 0.0030 H14 = 0.0021 H15 = 0.0016

H22 = 0.0044 H23 = 0.0026 H24 = 0.0018 H25 = 0.0014

H32 = 0.0038 H33 = 0.0022 H34 = 0.0016 H35 = 0.0012

H42 = 0.0031 H43 = 0.0019 H44 = 0.0013 H45 = 0.0010

H52 = 0.0025 H53 = 0.0015 H54 = 0.0011 H55 = 0.0008

H62 = 0.0019 H63 = 0.0011 H64 = 0.0008 H65 = 0.0006

H72 = 0.0012 H73 = 0.0007 H74 = 0.0005 H75 = 0.0004

H82 = 0.0006 H83 = 0.0004 H84 = 0.0003 H85 = 0.0002

H16 = 0.0013 H17 = 0.0011 H18 = 0.0009 H19 = 0.0008

H26 = 0.0011 H27 = 0.0009 H28 = 0.0008 H29 = 0.0007

H36 = 0.0010 H37 = 0.0008 H38 = 0.0007 H39 = 0.0006

H46 = 0.0008 H47 = 0.0007 H48 = 0.0006 H49 = 0.0005

H56 = 0.0007 H57 = 0.0005 H58 = 0.0005 H59 = 0.0004

H66 = 0.0005 H67 = 0.0004 H68 = 0.0004 H69 = 0.0003

H76 = 0.0003 H77 = 0.0003 H78 = 0.0002 H79 = 0.0002

H86 = 0.0002 H87 = 0.0001 H88 = 0.0001 H89 = 0.0001.

สำนกหอ


40

Thus cross sectional image of the ground structure of magnetic field is as follows.

Figure 4.2: The value of magnetic field when b = 0.001 m−1.

สำนกหอ


41

Graphs of the relationship between magnetic field and distance of receiver from

source at various depths.

(a)

(b)

Figure 4.3: The relationship between magnetic field and distance of receiver from

source at various depths as b = 0.001 m−1. (a) The value of magnetic field when

10 ≤ r ≤ 170 m and 0 ≤ z ≤ 160 m. (b) The value of magnetic field when z is

fixed, i.e. z = 0, 20, 40, . . . , 160 m (from top to bottom).

สำนกหอ


42

From Figure 4.3(a), when b = 0.001 m−1, we can see that the value of mag-

netic field decreases exponentially as r increases and it decreases as z increases.

From Figure 4.3(b), when b = 0.001 m−1, the value of magnetic field decreases

exponentially when r increases.

The values of magnetic field when b = 0.01 m−1 are calculated as

H12 = 0.0050 H13 = 0.0029 H14 = 0.0020 H15 = 0.0015

H22 = 0.0043 H23 = 0.0025 H24 = 0.0017 H25 = 0.0013

H32 = 0.0036 H33 = 0.0021 H34 = 0.0014 H35 = 0.0010

H42 = 0.0030 H43 = 0.0017 H44 = 0.0011 H45 = 0.0008

H52 = 0.0024 H53 = 0.0013 H54 = 0.0009 H55 = 0.0006

H62 = 0.0017 H63 = 0.0010 H64 = 0.0006 H65 = 0.0005

H72 = 0.0011 H73 = 0.0006 H74 = 0.0004 H75 = 0.0003

H82 = 0.0005 H83 = 0.0003 H84 = 0.0002 H85 = 0.0001

H16 = 0.0012 H17 = 0.0010 H18 = 0.0009 H19 = 0.0008

H26 = 0.0010 H27 = 0.0008 H28 = 0.0007 H29 = 0.0007

H36 = 0.0008 H37 = 0.0007 H38 = 0.0006 H39 = 0.0006

H46 = 0.0007 H47 = 0.0006 H48 = 0.0005 H49 = 0.0005

H56 = 0.0005 H57 = 0.0004 H58 = 0.0004 H59 = 0.0004

H66 = 0.0004 H67 = 0.0003 H68 = 0.0003 H69 = 0.0003

H76 = 0.0002 H77 = 0.0002 H78 = 0.0002 H79 = 0.0002

H86 = 0.0001 H87 = 0.0001 H88 = 0.0001 H89 = 0.0001.

สำนกหอ


43



สำนกหอ


44



(a)

(b)





สำนกหอ


45

From Figure 4.5(a), when b = 0.01 m−1, the behavior of magnetic field is

similar to that when b = 0.001 m−1. However, the value of magnetic field for the

case b = 0.01 m−1 is smaller than the case when b = 0.001 m−1. From Figure

4.5(b), when b = 0.01 m−1, the value of magnetic field decreases exponentially

when r increases and it is smaller than the case when b = 0.001 m−1.


H12 = 0.0045 H13 = 0.0022 H14 = 0.0014 H15 = 0.0009

H22 = 0.0036 H23 = 0.0018 H24 = 0.0010 H25 = 0.0006

H32 = 0.0030 H33 = 0.0014 H34 = 0.0007 H35 = 0.0004

H42 = 0.0024 H43 = 0.0011 H44 = 0.0005 H45 = 0.0003

H52 = 0.0019 H53 = 0.0008 H54 = 0.0004 H55 = 0.0002

H62 = 0.0013 H63 = 0.0005 H64 = 0.0002 H65 = 0.0001

H72 = 0.0016 H73 = 0.0003 H74 = 0.0001 H75 = 0.0001

H82 = 0.0008 H83 = 0.0001 H84 = 0.00005 H85 = 0.00002

H16 = 0.0007 H17 = 0.0006 H18 = 0.0006 H19 = 0.0006

H26 = 0.0004 H27 = 0.0004 H28 = 0.0004 H29 = 0.0005

H36 = 0.0003 H37 = 0.0003 H38 = 0.0003 H39 = 0.0004

H46 = 0.0002 H47 = 0.0002 H48 = 0.0002 H49 = 0.0003

H56 = 0.0001 H57 = 0.0001 H58 = 0.0002 H59 = 0.0002

H66 = 0.0001 H67 = 0.0001 H68 = 0.0001 H69 = 0.0002

H76 = 0.00004 H77 = 0.00004 H78 = 0.00006 H79 = 0.0001

H86 = 0.00001 H87 = 0.00001 H88 = 0.00002 H89 = 0.00004.

สำนกหอ


46



สำนกหอ


47



(a)

(b)





สำนกหอ


48

From Figure 4.7(a)-(b), when b = 0.05 m−1, the behavior of magnetic field

is similar to that when b = 0.001 m−1 and b = 0.01 m−1 but it increases slowly

again as r increases (we can see that r = 150 m and r = 170 m from above in

Figure 4.7(b)). However, the value of magnetic field for the case b = 0.05 m−1 is

smaller than cases when b = 0.001 m−1 and b = 0.01 m−1.


H12 = 0.0041 H13 = 0.0018 H14 = 0.0009 H15 = 0.0006

H22 = 0.0033 H23 = 0.0014 H24 = 0.0007 H25 = 0.0003

H32 = 0.0027 H33 = 0.0011 H34 = 0.0005 H35 = 0.0002

H42 = 0.0022 H43 = 0.0008 H44 = 0.0003 H45 = 0.0002

H52 = 0.0016 H53 = 0.0006 H54 = 0.0002 H55 = 0.0001

H62 = 0.0011 H63 = 0.0004 H64 = 0.0001 H65 = 0.0001

H72 = 0.0006 H73 = 0.0002 H74 = 0.0001 H75 = 0.00002

H82 = 0.0002 H83 = 0.0001 H84 = 0.00002 H85 = 0.00001

H16 = 0.0004 H17 = 0.0004 H18 = 0.0004 H19 = 0.0005

H26 = 0.0002 H27 = 0.0002 H28 = 0.0003 H29 = 0.0004

H36 = 0.0001 H37 = 0.0001 H38 = 0.0002 H39 = 0.0003

H46 = 0.0001 H47 = 0.0001 H48 = 0.0001 H49 = 0.0003

H56 = 0.0001 H57 = 0.0001 H58 = 0.0001 H59 = 0.0002

H66 = 0.00003 H67 = 0.00004 H68 = 0.0001 H69 = 0.0001

H76 = 0.00001 H77 = 0.00002 H78 = 0.00003 H79 = 0.0001

H86 = 0.000004 H87 = 0.000005 H88 = 0.00001 H89 = 0.00003.

สำนกหอ


49



สำนกหอ


50



(a)

(b)





สำนกหอ


51

From Figure 4.9(a)-(b), when b = 0.075 m−1, the behavior of magnetic

field is similar to that when b = 0.001, 0.01 and 0.05 m−1 but it increases slowly

again as r increases (we can see that r = 130, 150 and 170 m from above in Figure

4.9(b)). However, the value of magnetic field for the case b = 0.075 m−1 is smaller

than cases when b = 0.001, 0.01 and 0.05 m−1.


H12 = 0.0037 H13 = 0.0013 H14 = 0.0006 H15 = 0.0003

H22 = 0.0030 H23 = 0.0012 H24 = 0.0005 H25 = 0.0002

H32 = 0.0025 H33 = 0.0009 H34 = 0.0003 H35 = 0.0001

H42 = 0.0020 H43 = 0.0007 H44 = 0.0002 H45 = 0.0001

H52 = 0.0015 H53 = 0.0005 H54 = 0.0001 H55 = 0.00005

H62 = 0.0010 H63 = 0.0003 H64 = 0.0001 H65 = 0.00003

H72 = 0.0005 H73 = 0.0001 H74 = 0.00003 H75 = 0.00001

H82 = 0.0001 H83 = 0.00004 H84 = 0.00001 H85 = 0.000003

H16 = 0.0002 H17 = 0.0002 H18 = 0.0002 H19 = 0.0004

H26 = 0.0001 H27 = 0.0001 H28 = 0.0002 H29 = 0.0003

H36 = 0.0001 H37 = 0.0001 H38 = 0.0002 H39 = 0.0003

H46 = 0.00005 H47 = 0.00006 H48 = 0.0001 H49 = 0.0002

H56 = 0.00003 H57 = 0.00004 H58 = 0.0001 H59 = 0.0002

H66 = 0.00001 H67 = 0.00002 H68 = 0.00004 H69 = 0.0001

H76 = 0.000005 H77 = 0.00001 H78 = 0.00002 H79 = 0.0001

H86 = 0.000001 H87 = 0.000002 H88 = 0.00001 H89 = 0.00002.

สำนกหอ


52

The cross sectional image of the ground structure of magnetic field is as follows.


สำนกหอ


53



(a)

(b)





สำนกหอ


54


is similar to that when b = 0.001, 0.01, 0.05 and 0.075 m−1 but it increases slowly

again as r increases (we can see that r = 130, 150 and 170 m from above in Figure

4.11(b)). However, the value of magnetic field for the case b = 0.1 m−1 is smaller

than cases when b = 0.001, 0.01, 0.05 and 0.075 m−1.

สำนกหอ


55



(a) (b)

(c) (d)

(e)


source at various depths when (a) b = 0.001 m−1 (b) b = 0.01 m−1 (c) b = 0.05

m−1 (d) b = 0.075 m−1 (e) b = 0.1 m−1

สำนกหอ


56

From Figure 4.12(a) to (e), when b = 0.001, 0.01, 0.05, 0.075 and 0.1 m−1,

respectively, we can see that the value of magnetic field decreases exponentially as

r increases and it decreases as z increases. The value of magnetic field is highest

when b = 0.001 m−1 and it decreases when b increases.

สำนกหอ


57


source at various depths when z is fixed, i.e. z = 0, 20, 40, . . . , 160 m (from top to

bottom).

(a) (b)

(c) (d)

(e)


source when z is fixed and (a) b = 0.001 m−1 (b) b = 0.01 m−1 (c) b = 0.05 m−1

(d) b = 0.075 m−1 (e) b = 0.1 m−1

สำนกหอ


58

From Figure 4.13(a)−(e), when z is fixed, the value of magnetic field de-

creases exponentially as r increases. However, when b = 0.05, 0.075 and 0.1 m−1,

the value of magnetic field decreases exponentially as r increases and it increases

slowly again as r increases (see at r = 130, 150 and 170 m in Figure 4.13(c), (d)

and (e)).

สำนกหอ


59

(a) (b)

(c) (d)

(e)

Figure 4.14: The relationship between magnetic field and different depths when r

is fixed and (a) b = 0.001 m−1 (b) b = 0.01 m−1 (c) b = 0.05 m−1 (d) b = 0.075

m−1 (e) b = 0.1 m−1

สำนกหอ


60

Figure 4.14(a)-(e), represents the value of magnetic field when r is fixed

(10, 30, . . . , and 170) and b = 0.001, 0.01, 0.05, 0.075 and 0.1 m−1, respectively,

we can see that the value of magnetic field decreases as b increases. The top

line represents the value of magnetic field when r = 10 m and the bottom line

represents the value of magnetic field when r = 170 m. Figure 4.14(c),(d) and

(e) show a different behavior of magnetic field when b = 0.05, 0.075 and 0.1 m−1,

respectively. The value of magnetic field as r = 130, 150 and 170 m is greater than

that when r = 70, 90 and 110 m, as we can see the lines cross those three lines for

r = 70, 90 and 110 m.

สำนกหอ


61

(a) (b)

(c) (d)


source when b varies from 0.001, 0.01, 0.05, 0.075 and 0.1 m−1 and z is fixed. (a)

z=20 m (b) z=60 m (c) z=100 m (d) z=140 m

Figure 4.15(a) to (d), represents the value of magnetic field when b varies

and z is fixed at 20, 60, 100 and 140, respectively. We can see that the value of

magnetic field where b = 0.001, 0.01, 0.05, 0.075 and 0.1 m−1 decreases exponen-

tially as r increases and it has similar values when z increases because the value

of magnetic field decreases to zero and it has value near zero when z increases as

b varies.

สำนกหอ


62

Contour graphs of the relationship between magnetic field and distance of receiver

from source at various depths.

(a) (b)

(c) (d)

(e)

Figure 4.16: Contour graphs of magnetic field at different distances of receiver

from source and different depths when (a) b = 0.001 m−1 (b) b = 0.01 m−1 (c) b

= 0.05 m−1 (d) b = 0.075 m−1 (e) b = 0.1 m−1

สำนกหอ


63

From Figure 4.16(a) to (e), when b = 0.001, 0.01, 0.05, 0.075 and 0.1 m−1,

the red color shows the area when the value of magnetic field is high and the

blue color shows the area when the value of magnetic field is low. The value of

magnetic field decreases when b increases, as we can see in Figure 4.16(a) to (e).

4.2 Case of an Exponentially Increasing Conductivity

Turning to the case of increasing conductivity σ = σ0e−bz, where b < 0,

the numerical solutions of the magnetic field at each node is calculated by using

MATLAB program.

The values of magnetic field when b = −0.001 m−1 are computed as

H12 = 0.0052 H13 = 0.0030 H14 = 0.0021 H15 = 0.0016

H22 = 0.0044 H23 = 0.0027 H24 = 0.0019 H25 = 0.0014

H32 = 0.0038 H33 = 0.0023 H34 = 0.0016 H35 = 0.0012

H42 = 0.0032 H43 = 0.0019 H44 = 0.0014 H45 = 0.0010

H52 = 0.0025 H53 = 0.0015 H54 = 0.0011 H55 = 0.0008

H62 = 0.0019 H63 = 0.0012 H64 = 0.0008 H65 = 0.0006

H72 = 0.0013 H73 = 0.0008 H74 = 0.0006 H75 = 0.0004

H82 = 0.0006 H83 = 0.0004 H84 = 0.0003 H85 = 0.0002

H16 = 0.0013 H17 = 0.0011 H18 = 0.0009 H19 = 0.0008

H26 = 0.0012 H27 = 0.0010 H28 = 0.0008 H29 = 0.0007

H36 = 0.0010 H37 = 0.0008 H38 = 0.0007 H39 = 0.0006

H46 = 0.0008 H47 = 0.0007 H48 = 0.0006 H49 = 0.0005

H56 = 0.0007 H57 = 0.0006 H58 = 0.0005 H59 = 0.0004

H66 = 0.0005 H67 = 0.0004 H68 = 0.0004 H69 = 0.0003

H76 = 0.0003 H77 = 0.0003 H78 = 0.0003 H79 = 0.0002

H86 = 0.0002 H87 = 0.0001 H88 = 0.0001 H89 = 0.0001.

สำนกหอ


64

Thus cross sectional image of the ground structure of magnetic field is shown as

Figure 4.17: The value of magnetic field when b=-0.001 m−1.

สำนกหอ


65



(a)

(b)


source at various depths as b=-0.001 m−1. (a) The value of magnetic field when



สำนกหอ


66

From Figure 4.18(a), when b = −0.001 m−1, we can see that the value

of magnetic field decreases exponentially as r increases and it decreases as z in-

creases. From Figure 4.18(b), when b = −0.001 m−1, the value of magnetic field

decreases exponentially when r increases.

The values of magnetic field when b = −0.01 m−1 are calculated as

H12 = 0.0052 H13 = 0.0031 H14 = 0.0022 H15 = 0.0017

H22 = 0.0045 H23 = 0.0028 H24 = 0.0020 H25 = 0.0016

H32 = 0.0039 H33 = 0.0024 H34 = 0.0018 H35 = 0.0014

H42 = 0.0033 H43 = 0.0021 H44 = 0.0015 H45 = 0.0012

H52 = 0.0027 H53 = 0.0017 H54 = 0.0013 H55 = 0.0010

H62 = 0.0020 H63 = 0.0013 H64 = 0.0010 H65 = 0.0008

H72 = 0.0014 H73 = 0.0009 H74 = 0.0007 H75 = 0.0006

H82 = 0.0007 H83 = 0.0005 H84 = 0.0004 H85 = 0.0003

H16 = 0.0014 H17 = 0.0011 H18 = 0.0010 H19 = 0.0008

H26 = 0.0013 H27 = 0.0010 H28 = 0.0009 H29 = 0.0007

H36 = 0.0011 H37 = 0.0009 H38 = 0.0008 H39 = 0.0006

H46 = 0.0010 H47 = 0.0008 H48 = 0.0007 H49 = 0.0006

H56 = 0.0008 H57 = 0.0007 H58 = 0.0006 H59 = 0.0005

H66 = 0.0007 H67 = 0.0006 H68 = 0.0005 H69 = 0.0004

H76 = 0.0005 H77 = 0.0004 H78 = 0.0003 H79 = 0.0003

H86 = 0.0003 H87 = 0.0002 H88 = 0.0002 H89 = 0.0001.

สำนกหอ


67



สำนกหอ


68



(a)

(b)





สำนกหอ


69

From Figure 4.20(a), when b = −0.01 m−1, the behavior of magnetic field

is similar to that when b = −0.001 m−1. However, the value of magnetic field is

greater than the case when b = −0.001m−1. From Figure 4.20(b), when b = −0.01

m−1, the value of magnetic field decreases exponentially when r increases and it

is greater than the case when b = −0.001 m−1.


H12 = 0.0055 H13 = 0.0033 H14 = 0.0024 H15 = 0.0018

H22 = 0.0049 H23 = 0.0032 H24 = 0.0023 H25 = 0.0018

H32 = 0.0044 H33 = 0.0029 H34 = 0.0022 H35 = 0.0017

H42 = 0.0038 H43 = 0.0026 H44 = 0.0021 H45 = 0.0017

H52 = 0.0032 H53 = 0.0023 H54 = 0.0019 H55 = 0.0016

H62 = 0.0026 H63 = 0.0020 H64 = 0.0017 H65 = 0.0014

H72 = 0.0020 H73 = 0.0016 H74 = 0.0014 H75 = 0.0012

H82 = 0.0013 H83 = 0.0011 H84 = 0.0010 H85 = 0.0009

H16 = 0.0015 H17 = 0.0012 H18 = 0.0011 H19 = 0.0009

H26 = 0.0015 H27 = 0.0012 H28 = 0.0010 H29 = 0.0008

H36 = 0.0014 H37 = 0.0012 H38 = 0.0010 H39 = 0.0008

H46 = 0.0014 H47 = 0.0011 H48 = 0.0009 H49 = 0.0007

H56 = 0.0013 H57 = 0.0011 H58 = 0.0008 H59 = 0.0006

H66 = 0.0012 H67 = 0.0010 H68 = 0.0008 H69 = 0.0006

H76 = 0.0010 H77 = 0.0009 H78 = 0.0007 H79 = 0.0005

H86 = 0.0007 H87 = 0.0006 H88 = 0.0005 H89 = 0.0003.

สำนกหอ


70



สำนกหอ


71



(a)

(b)





สำนกหอ


72

From Figure 4.22(a)-(b), when b = −0.05 m−1, the behavior of magnetic

field is similar to that when b = −0.001 m−1 and b = −0.01 m−1. However, the

value of magnetic field for the case b = −0.05 m−1 is greater than cases when

b = −0.001 m−1 and b = −0.01 m−1.

The values of magnetic field when b = −0.075 m−1 are calculated as

H12 = 0.0056 H13 = 0.0034 H14 = 0.0024 H15 = 0.0018

H22 = 0.0051 H23 = 0.0033 H24 = 0.0024 H25 = 0.0018

H32 = 0.0046 H33 = 0.0031 H34 = 0.0023 H35 = 0.0018

H42 = 0.0040 H43 = 0.0028 H44 = 0.0022 H45 = 0.0018

H52 = 0.0035 H53 = 0.0026 H54 = 0.0021 H55 = 0.0017

H62 = 0.0029 H63 = 0.0023 H64 = 0.0019 H65 = 0.0016

H72 = 0.0023 H73 = 0.0020 H74 = 0.0017 H75 = 0.0015

H82 = 0.0016 H83 = 0.0015 H84 = 0.0014 H85 = 0.0012

H16 = 0.0015 H17 = 0.0012 H18 = 0.0011 H19 = 0.0009

H26 = 0.0015 H27 = 0.0012 H28 = 0.0010 H29 = 0.0008

H36 = 0.0015 H37 = 0.0012 H38 = 0.0010 H39 = 0.0008

H46 = 0.0014 H47 = 0.0012 H48 = 0.0010 H49 = 0.0007

H56 = 0.0014 H57 = 0.0012 H58 = 0.0009 H59 = 0.0007

H66 = 0.0014 H67 = 0.0011 H68 = 0.0009 H69 = 0.0006

H76 = 0.0013 H77 = 0.0010 H78 = 0.0008 H79 = 0.0005

H86 = 0.0011 H87 = 0.0009 H88 = 0.0007 H89 = 0.0004.

สำนกหอ


73



สำนกหอ


74



(a)

(b)





สำนกหอ


75


field is similar to that when b = −0.001,−0.01 and −0.05 m−1. However, the


b = −0.001,−0.01 and −0.05 m−1.


H12 = 0.0057 H13 = 0.0035 H14 = 0.0024 H15 = 0.0019

H22 = 0.0052 H23 = 0.0034 H24 = 0.0024 H25 = 0.0019

H32 = 0.0047 H33 = 0.0032 H34 = 0.0024 H35 = 0.0018

H42 = 0.0042 H43 = 0.0030 H44 = 0.0023 H45 = 0.0018

H52 = 0.0037 H53 = 0.0028 H54 = 0.0022 H55 = 0.0018

H62 = 0.0032 H63 = 0.0025 H64 = 0.0021 H65 = 0.0017

H72 = 0.0026 H73 = 0.0023 H74 = 0.0020 H75 = 0.0017

H82 = 0.0020 H83 = 0.0019 H84 = 0.0018 H85 = 0.0015

H16 = 0.0015 H17 = 0.0012 H18 = 0.0011 H19 = 0.0009

H26 = 0.0015 H27 = 0.0012 H28 = 0.0011 H29 = 0.0009

H36 = 0.0015 H37 = 0.0012 H38 = 0.0010 H39 = 0.0008

H46 = 0.0015 H47 = 0.0012 H48 = 0.0010 H49 = 0.0007

H56 = 0.0015 H57 = 0.0012 H58 = 0.0010 H59 = 0.0007

H66 = 0.0014 H67 = 0.0012 H68 = 0.0009 H69 = 0.0006

H76 = 0.0014 H77 = 0.0011 H78 = 0.0009 H79 = 0.0006

H86 = 0.0013 H87 = 0.0011 H88 = 0.0008 H89 = 0.0005.

สำนกหอ


76



สำนกหอ


77



(a)

(b)

Figure 4.26: The relationship between magnetic field and distance of receiver

from source at various depths as b=-0.1 m−1. (a) The value of magnetic field

when 10 ≤ r ≤ 170 m and 0 ≤ z ≤ 160 m. (b) The value of magnetic field when

z is fixed, i.e. z = 0, 20, 40, . . . , 160 m (from top to bottom).

สำนกหอ


78


field is similar to that when b = −0.001,−0.01,−0.05 and −0.075 m−1 but when

z = 160 m, it increases slowly as 10 < r < 30 m and decreases slowly again

as r increases (see at z = 160 m from above in Figure 4.26(b)). However, the


b = −0.001,−0.01,−0.05 and −0.075 m−1.

สำนกหอ


79



(a) (b)

(c) (d)

(e)


source at various depths when (a) b=-0.001m−1 (b) b=-0.01m−1 (c) b=-0.05m−1

(d) b=-0.075 m−1 (e) b=-0.1 m−1

สำนกหอ


80

From Figure 4.27(a) to (e), when b = −0.001,−0.01,−0.05,−0.075 and

−0.1 m−1, respectively, we can see that the value of magnetic field decreases

exponentially as r increases and it decreases as z increases as well. The value of

magnetic field increases when b decreases.

สำนกหอ


81



bottom).

(a) (b)

(c) (d)

(e)


source when z is fixed and (a) b=-0.001 m−1 (b) b=-0.01 m−1 (c) b=-0.05 m−1

(d) b=-0.075 m−1 (e) b=-0.1 m−1

สำนกหอ


82


creases exponentially as r increases. However, when b = −0.1 m−1, the value of

magnetic field increases slowly as 10 < r < 30 m and decreases slowly again as r

increases when z = 160 m (we can see in Figure 4.28(e)).

สำนกหอ


83

(a) (b)

(c) (d)

(e)

Figure 4.29: The relationship between magnetic field and different depths when

r is fixed and (a) b=-0.001 m−1 (b) b=-0.01 m−1 (c) b=-0.05 m−1 (d) b=-0.075

m−1 (e) b=-0.1 m−1

สำนกหอ


84

Figure 4.29(a)-(e), represents the value of magnetic field when r is fixed

(10, 30, . . . , and 170) and b = −0.001,−0.01,−0.05,−0.075 and −0.1 m−1, respec-

tively, we can see that the value of magnetic field increases as b decreases. The

top line represents the value of magnetic field when r = 10 m and the bottom line

represents the value of magnetic field when r = 170 m.

สำนกหอ


85

(a) (b)

(c) (d)

Figure 4.30: The relationship between magnetic field and distance of receiver

from source when b varies from −0.001,−0.01,−0.05,−0.075 and −0.1 m−1 and z

is fixed. (a) z=20 m (b) z=60 m (c) z=100 m (d) z=140 m

Figure 4.30(a) to (d), represents the value of magnetic field when b varies

and z is fixed at 20, 60, 100 and 140, respectively. We can see that the value of

magnetic field where b = −0.001,−0.01,−0.05,−0.075 and −0.1 m−1 decreases

exponentially as r increases and it has similar values when z increases. The value

of magnetic field increases when b decreases, as we can see in Figure 4.30(a) to

(d).

สำนกหอ


86



(a) (b)

(c) (d)

(e)

Figure 4.31: Contour graphs of magnetic field at different distances of receiver from

source and different depths when (a) b=-0.001 m−1 (b) b=-0.01 m−1 (c) b=-0.05

m−1 (d) b=-0.075 m−1 (e) b=-0.1 m−1

สำนกหอ


87

From Figure 4.31(a) to (e), when b = −0.001,−0.01,−0.05,−0.075 and

−0.1 m−1, the red color shows the area when the value of magnetic field is high

and the blue color shows the area when the value of magnetic field is low. The value

of magnetic field increases when b decreases, as we can see in Figure 4.31(a) to (e).

Summarize

In this chapter, finite element method is used to approximate the solution

of partial differential equation. The Maxwell’s equation is our governing equation

that can be used to find magnetic field. Under the boundary conditions and the

conductivity of the ground as σ = σ0e−bz we obtain the behavior of magnetic field

decreases to zero when the depth of soil increases. As well as the case of increasing

the space between source-receiver, the magnetic field decreases to zero too. The

value of b is an important role for the conduction of ground and effect to the

magnetic field quantities as well. The comparision of the quantities of magnetic

field for the case of σ as an increasing function is higher than the case of σ as a

decreasing function according to the advantage of the DC source that better reflex

at very large depth than on the ground surface.

Since we used the exponential function of the subsurface as given by σ =

σ0e−bz, where σ0 > 0 and b may be either positive or negative. In case of an

exponentially increasing ground profile, b < 0, we obtain σ → ∞ as z → ∞. In

case of an exponentially decreasing ground profile, b > 0, we obtain σ → 0 as z →∞. Then, we use the conductivity developed of Yooyuanyong and Chumchob[37]

that a more realistic model as σ(z) = σ0 + (σ1 − σ0)e−bz, where σ0, σ1 and b are

positive constants. In case of an exponentially increasing ground profile, σ0 > σ1,

we obtain σ → σ0 as z → ∞. In case of an exponentially decreasing ground

profile, σ0 < σ1, we obtain σ → σ0 as z → ∞, as shown in chapter 5.

สำนกหอ


Chapter 5

Numerical Results for the case of

σ(z) = σ0 + (σ1 − σ0)e−bz

In this chapter, we consider numerical solution of problem having expo-

nential conductivity σ(z) = σ0 + (σ1 − σ0)e−bz, where σ0, σ1 and b are positive

constants.

Figure 5.1: Geometric model of the Earth structure in the case of σ(z) = σ0 +

(σ1 − σ0)e−bz.

Substituing σ(z) = σ0 + (σ1 − σ0)e−bz into equation (3.15), we obtain

100∑j=1

Hj

[−

∫Ω


∫Ω

br

((σ1 − σ0)e

−bz

σ0 + (σ1 − σ0)e−bz

)∂ϕj

∂zϕidrdz

−∫Ω

1

rϕjϕidrdz

]= 0, (5.1)

88

สำนกหอ


89

for each i = 1, 2, . . . , 100 and

Mi,j = −∫Ω


∫Ω

br

((σ1 − σ0)e

−bz

σ0 + (σ1 − σ0)e−bz

)∂ϕj

∂zϕidrdz

−∫Ω

1

rϕjϕidrdz.

Next, we consider the value of Mi,j. We separate Mi,j into 3 terms, i.e.

Mi,j = Ai,j + Bi,j + Ci,j where

Ai,j = −∫Ω

r∇ϕj · ∇ϕidrdz, Bi,j =

∫Ω

br

((σ1 − σ0)e

−bz

σ0 + (σ1 − σ0)e−bz

)∂ϕj

∂zϕidrdz, Ci,j =

−∫Ω

1

rϕjϕidrdz.

To calculate elements in Ai,j, Bi,j and Ci,j, we transform r, z to ξ, η, re-

spectively, by using the transformation equation (3.16) together with basis func-

tions equation (3.18). We can see the value of Ai,j and Ci,j from the case of

σ(z) = σ0e−bz, because they are the same.

Consider the value of Bi,j =

∫Ω

br

((σ1 − σ0)e

−bz

σ0 + (σ1 − σ0)e−bz

)∂ϕj

∂zϕidrdz by

using Chain rule and Jacobian transform thus

∫Ω

br

((σ1 − σ0)e

−bz

σ0 + (σ1 − σ0)e−bz

)∂ϕj

∂zϕidrdz

=

∫ 1

−1

∫ 1

−1

br

((σ1 − σ0)e

−bz

σ0 + (σ1 − σ0)e−bz

)[ ˆ∂ϕj

∂ξ

∂ξ

∂z+

ˆ∂ϕj

∂η

∂η

∂z

]ϕi

(h2

)2

dξdη

=

∫ 1

−1

∫ 1

−1

br

((σ1 − σ0)e

−bz

σ0 + (σ1 − σ0)e−bz

)[ ˆ∂ϕj

∂ξ0 +

ˆ∂ϕj

∂η

2

h

]ϕi

(h2

)2

dξdη

=bh

2

∫ 1

−1

∫ 1

−1

r

((σ1 − σ0)e

−bz

σ0 + (σ1 − σ0)e−bz

) ˆ∂ϕj

∂ηϕidξdη

=bh

2

∫ 1

−1

∫ 1

−1

(rk +

h

2(1 + ξ)

)( (σ1 − σ0)e−b(zk+

h2(1+η)

)σ0 + (σ1 − σ0)e

−b(zk+

h2(1+η)

))

ˆ∂ϕj

∂ηϕidξdη,

where k = 0, 1, . . . , 8 and ϕi, ϕj are basis function in coordinate ξ, η for each

i, j = 1, 2, . . . , 100.

For the corresponding linear rectangular elements, the approximation

bh

2

∫ 1

−1

∫ 1

−1

(rk+

h

2(1+ξ)

)( (σ1 − σ0)e−b(zk+

h2(1+η)

)σ0 + (σ1 − σ0)e

−b(zk+

h2(1+η)

))

ˆ∂ϕj

∂ηϕidξdη can be di-

vided into nine cases (Figure. 3.3(b)).

สำนกหอ


90Tab

le5.1:

Thevalueof

bh 2

∫ 1 −1

∫ 1 −1

( r k+

h 2(1

+ξ))(

(σ1−

σ0)e

−b( z k

+h 2(1+η))

σ0+(σ

1−

σ0)e

−b( z k

+h 2(1+η))) ˆ

∂ϕj

∂ηϕidξdη

incoordinates

(ξ,η)

bh 2

∫ 1 −1

∫ 1 −1

( r k+

h 2(1

+ξ))(

(σ1−

σ0)e

−b( z

k+

h 2(1

+η))

σ0+

(σ1−

σ0)e

−b( z

k+

h 2(1

+η)))

ˆ∂ϕj

∂η

ϕidξdη

Elemen

tsCases

inco

ord

inates(ξ,η

)Results

Solutions

bh 2

∫ 1 −1

∫ 1 −1T1

((σ

1−

σ0)e

−b( z

k+

1+

h 2(1

+η))

σ0+

(σ1−

σ0)e

−b( z

k+

1+

h 2(1

+η))) (

∂N

3

∂η

N3

) dξdη

+bh 2

∫ 1 −1

∫ 1 −1T2

((σ

1−

σ0)e

−b( z

k+

1+

h 2(1

+η))

σ0+

(σ1−

σ0)e

−b( z

k+

1+

h 2(1

+η))) (

∂N

4

∂η

N4

) dξdη

+bh 2

∫ 1 −1

∫ 1 −1T1

((σ

1−

σ0)e

−b( z

k+

h 2(1

+η))

σ0+

(σ1−

σ0)e

−b( z

k+

h 2(1

+η))) (

∂N

2

∂η

N2

) dξdη

i=j

+bh 2

∫ 1 −1

∫ 1 −1T2

((σ

1−

σ0)e

−b( z

k+

h 2(1

+η))

σ0+

(σ1−

σ0)e

−b( z

k+

h 2(1

+η))) (

∂N

1

∂η

N1

) dξdη

h32bW

1

( mk+1

) +h

32bW

1

( nk+1

)h

32bW

1

( mk+1+

nk+1

)

bh 2

∫ 1 −1

∫ 1 −1T2

((σ

1−

σ0)e

−b( z

k+

1+

h 2(1

+η))

σ0+

(σ1−

σ0)e

−b( z

k+

1+

h 2(1

+η))) (

∂N

4

∂η

N3

) dξdη

i=j+

1+bh 2

∫ 1 −1

∫ 1 −1T2

((σ

1−

σ0)e

−b( z

k+

h 2(1

+η))

σ0+

(σ1−

σ0)e

−b( z

k+

h 2(1

+η))) (

∂N

1

∂η

N2

) dξdη

h32bW

3

( mk+1+

nk+1

)h

32bW

3

( mk+1+

nk+1

)

bh 2

∫ 1 −1

∫ 1 −1T1

((σ

1−

σ0)e

−b( z

k+

1+

h 2(1

+η))

σ0+

(σ1−

σ0)e

−b( z

k+

1+

h 2(1

+η))) (

∂N

3

∂η

N4

) dξdη

i=j-1

+bh 2

∫ 1 −1

∫ 1 −1T1

((σ

1−

σ0)e

−b( z

k+

h 2(1

+η))

σ0+

(σ1−

σ0)e

−b( z

k+

h 2(1

+η))) (

∂N

2

∂η

N1

) dξdη

h32bW

2

( mk+1+

nk+1

)h

32bW

2

( mk+1+

nk+1

)

bh 2

∫ 1 −1

∫ 1 −1T1

((σ

1−

σ0)e

−b( z

k+

1+

h 2(1

+η))

σ0+

(σ1−

σ0)e

−b( z

k+

1+

h 2(1

+η))) (

∂N

3

∂η

N2

) dξdη

i=j+

M+bh 2

∫ 1 −1

∫ 1 −1T2

((σ

1−

σ0)e

−b( z

k+

1+

h 2(1

+η))

σ0+

(σ1−

σ0)e

−b( z

k+

1+

h 2(1

+η))) (

∂N

4

∂η

N1

) dξdη

h32bW

1

( r k+1

)h

32bW

1

( r k+1

)

สำนกหอ


91

bh

2

∫ 1

−1

∫ 1

−1

(rk +

h

2(1 + ξ)

)( (σ1 − σ0)e−b

(zk+

h2(1+η)

)σ0 + (σ1 − σ0)e

−b(zk+

h2(1+η)

))

ˆ∂ϕj

∂ηϕidξdη

Elements Cases in coordinates (ξ, η) Results Solutions

bh

2

∫ 1

−1

∫ 1

−1T1

((σ1 − σ0)e

−b(zk+

h2(1+η)

)σ0 + (σ1 − σ0)e

−b(zk+

h2(1+η)

))(

∂N2

∂ηN3

)dξdη

i=j-Mbh

2

∫ 1

−1

∫ 1

−1T2

((σ1 − σ0)e

−b(zk+

h2(1+η)

)σ0 + (σ1 − σ0)e

−b(zk+

h2(1+η)

))(

∂N1

∂ηN4

)dξdη h

32bW1sk+1

h32b

W1

(sk+1

)

i=j+M+1bh

2

∫ 1

−1

∫ 1

−1T2

((σ1 − σ0)e

−b(zk+1+

h2(1+η)

)σ0 + (σ1 − σ0)e

−b(zk+1+

h2(1+η)

))(

∂N4

∂ηN2

)dξdη h

32bW3rk+1

h32b

W3

(rk+1

)

i=j-M-1bh

2

∫ 1

−1

∫ 1

−1T1

((σ1 − σ0)e

−b(zk+

h2(1+η)

)σ0 + (σ1 − σ0)e

−b(zk+

h2(1+η)

))(

∂N2

∂ηN4

)dξdη h

32bW2sk+1

h32b

W2

(sk+1

)

i=j+M-1bh

2

∫ 1

−1

∫ 1

−1T1

((σ1 − σ0)e

−b(zk+1+

h2(1+η)

)σ0 + (σ1 − σ0)e

−b(zk+1+

h2(1+η)

))(

∂N3

∂ηN1

)dξdη h

32bW2rk+1

h32b

W2

(rk+1

)

i=j-M+1bh

2

∫ 1

−1

∫ 1

−1T2

((σ1 − σ0)e

−b(zk+

h2(1+η)

)σ0 + (σ1 − σ0)e

−b(zk+

h2(1+η)

))(

∂N1

∂ηN3

)dξdη h

32bW3sk+1

h32b

W3

(sk+1

)

สำนกหอ


92

where k = 0, 1, . . . , 8 for each i, j = 1, 2, . . . , 100,

mk+1 =

[− 20(k + 1) ln

((σ1 + σ0e

20(k+1)b − σ0)e−20(k+1)b

)b

+20(k + 1) ln(

(σ1+σ0e20(k+1)b−σ0)e−20(k+1)b

σ0

)b

−dilog(

(σ1+σ0e20(k+1)b−σ0)e−20(k+1)b

σ0

)+20(k + 1) ln

((σ1 + σ0e

(20+20(k+1))b − σ0)e−(20+20(k+1))b

)b

−(20 + 20(k + 1)) ln(

(σ1+σ0e(20+20(k+1))b−σ0)e−(20+20(k+1))b

σ0

)+dilog

((σ1+σ0e(20+20(k+1))b−σ0)e−(20+20(k+1))b

σ0

)],

nk+1 =

[− 20(k + 1) ln

((σ1 + σ0e

20(k)b − σ0)e−20(k)b

)b

+20(k) ln(

(σ1+σ0e20(k)b−σ0)e−20(k)b

σ0

)b− dilog

((σ1+σ0e20(k)b−σ0)e−20(k)b

σ0

)+20(k + 1) ln

((σ1 + σ0e

20(k+1)b − σ0)e−20(k+1)b

)b

−20(k + 1) ln(

(σ1+σ0e20(k+1)b−σ0)e−20(k+1)b

σ0

)+dilog

((σ1+σ0e20(k+1)b−σ0)e−20(k+1)b

σ0

)],

rk+1 =

[(20 + 20(k + 1)) ln

((σ1 + σ0e

20(k+1)b − σ0)e−20(k+1)b

)b

−20(k + 1) ln(

(σ1+σ0e20(k+1)b−σ0)e−20(k+1)b

σ0

)b

+dilog(

(σ1+σ0e20(k+1)b−σ0)e−20(k+1)b

σ0

)−(20 + 20(k + 1)) ln

((σ1 + σ0e

(20+20(k+1))b − σ0)e−(20+20(k+1))b

)b

+(20 + 20(k + 1)) ln(

(σ1+σ0e(20+20(k+1))b−σ0)e−(20+20(k+1))b

σ0

)−dilog

((σ1+σ0e(20+20(k+1))b−σ0)e−(20+20(k+1))b

σ0

)],

sk+1 =

[20(k) ln

((σ1 + σ0e

20(k)b − σ0)e−20(k)b

)b

−20(k) ln(

(σ1+σ0e20(k)b−σ0)e−20(k)b

σ0

)b+ dilog

((σ1+σ0e20(k)b−σ0)e−20(k)b

σ0

)−20(k) ln

((σ1 + σ0e

20(k+1)b − σ0)e−20(k+1)b

)b

+20(k + 1) ln(

(σ1+σ0e20(k+1)b−σ0)e−20(k+1)b

σ0

)−dilog

((σ1+σ0e20(k+1)b−σ0)e−20(k+1)b

σ0

)],

W1 = ek =(

480+320k300

), W2 = fk =

(80(k+1)

300

), W3 = gk =

(80(k+2)

300

)and T1 =(

rk + h2(1 + ξ)

), T2 =

(rk+1 +

h2(1 + ξ)

)and dilog(x) =

∫ x

1

ln t

1− tdt for each

k = 0, 1, . . . , 8.

สำนกหอ


93

Then, from the value of

bh

2

∫ 1

−1

∫ 1

−1

(rk+

h

2(1+ξ)

)( (σ1 − σ0)e−b(zk+

h2(1+η)

)σ0 + (σ1 − σ0)e

−b(zk+

h2(1+η)

))

ˆ∂ϕj

∂ηϕidξdη can be writ-

ten in the form of matrix as follows

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

B1 B9 0 0 0 0 0 0

B16 B2 B10 0 0 0 0 0

0 B17 B3 B11 0 0 0 0

0 0 B18 B4 B12 0 0 0

0 0 0 B19 B5 B13 0 0

0 0 0 0 B20 B6 B14 0

0 0 0 0 0 B21 B7 B15

0 0 0 0 0 0 B22 B8

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

u1

u2

u3

u4

u5

u6

u7

u8

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

=

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

d1

d2

d3

d4

d5

d6

d7

d8

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

or BU = Q,

where

B =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

B1 B9 0 0 0 0 0 0

B16 B2 B10 0 0 0 0 0

0 B17 B3 B11 0 0 0 0

0 0 B18 B4 B12 0 0 0

0 0 0 B19 B5 B13 0 0

0 0 0 0 B20 B6 B14 0

0 0 0 0 0 B21 B7 B15

0 0 0 0 0 0 B22 B8

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦


]T, Q =

[d1, . . . , d8

]Tfor all i = 1, 2, . . . , 8 and

B1 =h32b

⎡⎢⎢⎢⎣

e0(m1+n1) g0(m1+n1) 0 0 0 0 0 0f1(m1+n1) e1(m1+n1) g1(m1+n1) 0 0 0 0 0

0 f2(m1+n1) e2(m1+n1) g2(m1+n1) 0 0 0 00 0 f3(m1+n1) e3(m1+n1) g3(m1+n1) 0 0 00 0 0 f4(m1+n1) e4(m1+n1) g4(m1+n1) 0 00 0 0 0 f5(m1+n1) e5(m1+n1) g5(m1+n1) 00 0 0 0 0 f6(m1+n1) e6(m1+n1) g6(m1+n1)0 0 0 0 0 0 f7(m1+n1) e7(m1+n1)

⎤⎥⎥⎥⎦,

สำนกหอ


94

B2 =h32b

⎡⎢⎢⎢⎣



⎤⎥⎥⎥⎦,

B3 =h32b

⎡⎢⎢⎢⎣



⎤⎥⎥⎥⎦,

B4 =h32b

⎡⎢⎢⎢⎣

e0(m4+n4) g0(m4+n4) 0 0 0 0 0 0f1(m4+n4) e1(m4+n4)) g1(m4+n4) 0 0 0 0 0


⎤⎥⎥⎥⎦,

B5 =h32b

⎡⎢⎢⎢⎣



⎤⎥⎥⎥⎦,

B6 =h32b

⎡⎢⎢⎢⎣



⎤⎥⎥⎥⎦,

B7 =h32b

⎡⎢⎢⎢⎣



⎤⎥⎥⎥⎦,

B8 =h32b

⎡⎢⎢⎢⎣



⎤⎥⎥⎥⎦,

สำนกหอ


95

B9 =h32b

⎡⎢⎢⎢⎢⎣

e0(r1) g0(r1) 0 0 0 0 0 0f1(r1) e1(r1) g1(r1) 0 0 0 0 0

0 f2(r1) e2(r1) g2(r1) 0 0 0 00 0 f3(r1) e3(r1) g3(r1) 0 0 00 0 0 f4(r1) e4(r1) g4(r1) 0 00 0 0 0 f5(r1) e5(r1) g5(r1) 00 0 0 0 0 f6(r1) e6(r1) g6(r1)0 0 0 0 0 0 f7(r1) e7(r1)

⎤⎥⎥⎥⎥⎦,

B10 =h32b

⎡⎢⎢⎢⎢⎣

e0(r2) g0(r2) 0 0 0 0 0 0f1(r2) e1(r2) g1(r2) 0 0 0 0 0


⎤⎥⎥⎥⎥⎦,

B11 =h32b

⎡⎢⎢⎢⎢⎣

e0(r3) g0(r3) 0 0 0 0 0 0f1(r3) e1(r3) g1(r3) 0 0 0 0 0


⎤⎥⎥⎥⎥⎦,

B12 =h32b

⎡⎢⎢⎢⎢⎣

e0(r4) g0(r4) 0 0 0 0 0 0f1(r4) e1(r4) g1(r4) 0 0 0 0 0


⎤⎥⎥⎥⎥⎦,

B13 =h32b

⎡⎢⎢⎢⎢⎣

e0(r5) g0(r5) 0 0 0 0 0 0f1(r5) e1(r5) g1(r5) 0 0 0 0 0


⎤⎥⎥⎥⎥⎦,

B14 =h32b

⎡⎢⎢⎢⎢⎣

e0(r6) g0(r6) 0 0 0 0 0 0f1(r6) e1(r6) g1(r6) 0 0 0 0 0


⎤⎥⎥⎥⎥⎦,

B15 =h32b

⎡⎢⎢⎢⎢⎣

e0(r7) g0(r7) 0 0 0 0 0 0f1(r7) e1(r7) g1(r7) 0 0 0 0 0


⎤⎥⎥⎥⎥⎦,

สำนกหอ


96

B16 =h32b

⎡⎢⎢⎢⎢⎣

e0(s2) g0(s2) 0 0 0 0 0 0f1(s2) e1(s2) g1(s2) 0 0 0 0 0

0 f2(s2) e2(s2) g2(s2) 0 0 0 00 0 f3(s2) e3(s2) g3(s2) 0 0 00 0 0 f4(s2) e4(s2) g4(s2) 0 00 0 0 0 f5(s2) e5(s2) g5(s2) 00 0 0 0 0 f6(s2) e6(s2) g6(s2)0 0 0 0 0 0 f7(s2) e7(s2)

⎤⎥⎥⎥⎥⎦,

B17 =h32b

⎡⎢⎢⎢⎢⎣

e0(s3) g0(s3) 0 0 0 0 0 0f1(s3) e1(s3) g1(s3) 0 0 0 0 0


⎤⎥⎥⎥⎥⎦,

B18 =h32b

⎡⎢⎢⎢⎢⎣

e0(s4) g0(s4) 0 0 0 0 0 0f1(s4) e1(s4) g1(s4) 0 0 0 0 0


⎤⎥⎥⎥⎥⎦,

B19 =h32b

⎡⎢⎢⎢⎢⎣

e0(s5) g0(s5) 0 0 0 0 0 0f1(s5) e1(s5) g1(s5) 0 0 0 0 0


⎤⎥⎥⎥⎥⎦,

B20 =h32b

⎡⎢⎢⎢⎢⎣

e0(s6) g0(s6) 0 0 0 0 0 0f1(s6) e1(s6) g1(s6) 0 0 0 0 0


⎤⎥⎥⎥⎥⎦,

B21 =h32b

⎡⎢⎢⎢⎢⎣

e0(s7) g0(s7) 0 0 0 0 0 0f1(s7) e1(s7) g1(s7) 0 0 0 0 0


⎤⎥⎥⎥⎥⎦,

B22 =h32b

⎡⎢⎢⎢⎢⎣

e0(s8) g0(s8) 0 0 0 0 0 0f1(s8) e1(s8) g1(s8) 0 0 0 0 0


⎤⎥⎥⎥⎥⎦,

สำนกหอ


97

d1 =bh96

⎡⎢⎢⎢⎢⎣

f0(s1)(0.0159)+e0(s1)(0.0053)+g0(s1)(0.0032)+f0(m1+n1)(0.0141)+f0(r1)(0.0124)f1(s1)(0.0053)+e1(s1)(0.0032)+g1(s1)(0.0023)f2(s1)(0.0032)+e2(s1)(0.0023)+g2(s1)(0.0018)f3(s1)(0.0023)+e3(s1)(0.0018)+g3(s1)(0.0014)f4(s1)(0.0018)+e4(s1)(0.0014)+g4(s1(0.0012)f5(s1)(0.0014)+e5(s1)(0.0012)+g5(s1(0.0011)f6(s1)(0.0012)+e6(s1)(0.0011)+g6(s1(0.0009)

f7(s1)(0.0011)+e7(s1)(0.0009)+g7(s1)(0.0008)+g7(m1+n1)(0.0007)+g7(r1)(0.0006)

⎤⎥⎥⎥⎥⎦,

d2 =bh96

⎡⎢⎢⎣

f0(s2)(0.0141)+f0(m2+n2)(0.0124)+f0(r2)(0.0106)000000

g7(s2)(0.0007)+g7(m2+n2)(0.0006)+g7(r2)(0.0005)

⎤⎥⎥⎦ ,

d3 =bh96

⎡⎢⎢⎣

f0(s3)(0.0124)+f0(m3+n3)(0.0106)+f0(r3)(0.0088)000000

g7(s3)(0.0006)+g7(m3+n3)(0.0005)+g7(r3)(0.00045)

⎤⎥⎥⎦,

d4 =bh96

⎡⎢⎢⎣

f0(s4)(0.0106)+f0(m4+n4)(0.0088)+f0(r4)(0.0071)000000

g7(s4)(0.0005)+g7(m4+n4)(0.00045)+g7(r4)(0.0004)

⎤⎥⎥⎦ ,

d5 =bh96

⎡⎢⎢⎣

f0(s5)(0.0088)+f0(m5+n5)(0.0071)+f0(r5)(0.0053)000000

g7(s5)(0.00045)+g7(m5+n5)(0.0004)+g7(r5)(0.0003)

⎤⎥⎥⎦,

d6 =bh96

⎡⎢⎢⎣

f0(s6)(0.0071)+f0(m6+n6)(0.0053)+f0(r6)(0.0035)000000

g7(s6)(0.0004)+g7(m6+n6)(0.0003)+g7(r6)(0.0002)

⎤⎥⎥⎦ ,

d7 =bh96

⎡⎢⎢⎣

f0(s7)(0.0053)+f0(m7+n7)(0.0035)+f0(r7)(0.0018)000000

g7(s7)(0.0003)+g7(m7+n7)(0.0002)+g7(r7)(0.0001)

⎤⎥⎥⎦,

d8 =bh96

⎡⎢⎢⎣

f0(s8)(0.0035)+f0(m8+n8)(0.0018)000000

g7(s8)(0.0002)+g7(m8+n8)(0.0001)

⎤⎥⎥⎦.

สำนกหอ


98

Therefore equation (5.1) can be written in the form of matrix as follows

(−A+ B − C

)

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

u1

u2

u3

u4

u5

u6

u7

u8

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

=

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

a1 + d1 + c1

a2 + d2 + c2

a3 + d3 + c3

a4 + d4 + c4

a5 + d5 + c5

a6 + d6 + c6

a7 + d7 + c7

a8 + d8 + c8

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

or(−A+ B − C

)ui =

(a+ d+ c

),

where ui =[H10i+2 H10i+3 H10i+4 H10i+5 H10i+6 H10i+7 H10i+8 H10i+9

]T,

P =[a1, . . . , a8

]T, Q =

[d1, . . . , d8

]Tand R =

[c1, . . . , c8

]Tfor all i = 1, 2, . . . , 8.

5.1 Case of an Exponentially Decreasing Conductivity

Applying the Galerkin’s Method of Weighted Residuals to equation (3.7),

we obtained the values of magnetic field at various positions of the earth’s structure

with one layer having exponentially decreasing conductivity σ = σ0+(σ1−σ0)e−bz

when σ0 < σ1. There is a source providing a DC voltage and a receiver on the

ground surface which picks up the signal from r = 10 m to r = 190 m. We

discrete the depth into 9 subintervals equally of the size h = 20 m, i.e. we

consider z = 0, 20, . . . , 180 m. We use constant σ1 = 1.5 S/m, σ0 = 0.5 S/m and

b = 0.01, 0.05, 0.1, 0.2 and 0.3 m−1. The numerical solutions of the magnetic field

at each node is calculated by using MATLAB program.

สำนกหอ


99


H12 = 0.0051 H13 = 0.0029 H14 = 0.0020 H15 = 0.0015

H22 = 0.0043 H23 = 0.0025 H24 = 0.0017 H25 = 0.0013

H32 = 0.0036 H33 = 0.0021 H34 = 0.0014 H35 = 0.0011

H42 = 0.0030 H43 = 0.0017 H44 = 0.0012 H45 = 0.0009

H52 = 0.0024 H53 = 0.0014 H54 = 0.0009 H55 = 0.0007

H62 = 0.0018 H63 = 0.0010 H64 = 0.0007 H65 = 0.0005

H72 = 0.0012 H73 = 0.0007 H74 = 0.0004 H75 = 0.0003

H82 = 0.0006 H83 = 0.0003 H84 = 0.0002 H85 = 0.0002

H16 = 0.0012 H17 = 0.0010 H18 = 0.0009 H19 = 0.0008

H26 = 0.0010 H27 = 0.0009 H28 = 0.0007 H29 = 0.0007

H36 = 0.0009 H37 = 0.0007 H38 = 0.0006 H39 = 0.0006

H46 = 0.0007 H47 = 0.0006 H48 = 0.0005 H49 = 0.0005

H56 = 0.0005 H57 = 0.0005 H58 = 0.0004 H59 = 0.0004

H66 = 0.0004 H67 = 0.0003 H68 = 0.0003 H69 = 0.0003

H76 = 0.0003 H77 = 0.0002 H78 = 0.0002 H79 = 0.0002

H86 = 0.0001 H87 = 0.0001 H88 = 0.0001 H89 = 0.0001.

สำนกหอ


100



สำนกหอ


101



(a)

(b)





สำนกหอ


102






H12 = 0.0042 H13 = 0.0022 H14 = 0.0014 H15 = 0.0010

H22 = 0.0035 H23 = 0.0017 H24 = 0.0010 H25 = 0.0007

H32 = 0.0031 H33 = 0.0015 H34 = 0.0009 H35 = 0.0006

H42 = 0.0027 H43 = 0.0014 H44 = 0.0008 H45 = 0.0005

H52 = 0.0023 H53 = 0.0012 H54 = 0.0007 H55 = 0.0005

H62 = 0.0018 H63 = 0.0009 H64 = 0.0006 H65 = 0.0004

H72 = 0.0012 H73 = 0.0006 H74 = 0.0004 H75 = 0.0003

H82 = 0.0006 H83 = 0.0003 H84 = 0.0002 H85 = 0.0001

H16 = 0.0008 H17 = 0.0007 H18 = 0.0006 H19 = 0.0006

H26 = 0.0005 H27 = 0.0005 H28 = 0.0004 H29 = 0.0005

H36 = 0.0005 H37 = 0.0004 H38 = 0.0004 H39 = 0.0004

H46 = 0.0004 H47 = 0.0004 H48 = 0.0003 H49 = 0.0004

H56 = 0.0004 H57 = 0.0003 H58 = 0.0003 H59 = 0.0003

H66 = 0.0003 H67 = 0.0003 H68 = 0.0003 H69 = 0.0003

H76 = 0.0002 H77 = 0.0002 H78 = 0.0002 H79 = 0.0002

H86 = 0.0001 H87 = 0.0001 H88 = 0.0001 H89 = 0.0001.

สำนกหอ


103



สำนกหอ


104



(a)

(b)





สำนกหอ


105


similar to that when b = 0.01 m−1. However, the value of magnetic field is smaller

than the case when b = 0.01 m−1. From Figure 5.5(b), when b = 0.05 m−1, the

value of magnetic field decreases exponentially when r increases and it is smaller

than the case when b = 0.01 m−1.


H12 = 0.0034 H13 = 0.0016 H14 = 0.0010 H15 = 0.0007

H22 = 0.0032 H23 = 0.0014 H24 = 0.0008 H25 = 0.0005

H32 = 0.0032 H33 = 0.0015 H34 = 0.0008 H35 = 0.0006

H42 = 0.0028 H43 = 0.0014 H44 = 0.0008 H45 = 0.0006

H52 = 0.0023 H53 = 0.0012 H54 = 0.0008 H55 = 0.0005

H62 = 0.0018 H63 = 0.0010 H64 = 0.0006 H65 = 0.0004

H72 = 0.0012 H73 = 0.0007 H74 = 0.0004 H75 = 0.0003

H82 = 0.0006 H83 = 0.0003 H84 = 0.0002 H85 = 0.0002

H16 = 0.0006 H17 = 0.0005 H18 = 0.0005 H19 = 0.0005

H26 = 0.0004 H27 = 0.0003 H28 = 0.0003 H29 = 0.0004

H36 = 0.0004 H37 = 0.0004 H38 = 0.0004 H39 = 0.0004

H46 = 0.0004 H47 = 0.0004 H48 = 0.0004 H49 = 0.0004

H56 = 0.0004 H57 = 0.0003 H58 = 0.0003 H59 = 0.0003

H66 = 0.0003 H67 = 0.0003 H68 = 0.0003 H69 = 0.0003

H76 = 0.0002 H77 = 0.0002 H78 = 0.0002 H79 = 0.0002

H86 = 0.0001 H87 = 0.0001 H88 = 0.0001 H89 = 0.0001.

สำนกหอ


106



สำนกหอ


107



(a)

(b)





สำนกหอ


108

From Figure 5.7(a)-(b), when b = 0.1 m−1, the behavior of magnetic

field is similar to that when b = 0.01 m−1 and b = 0.05 m−1 but it decreases

rapidly and increases slowly as r increases when z = 40 m (see the second line

from above in Figure 5.7(b)). However, the value of magnetic field for the case

b = 0.1 m−1 is smaller than the case when b = 0.01 m−1 and b = 0.05 m−1.

For z = 0, 20, 60, 80, 100, 120, 140 and 160 m, and so on, the value of magnetic

field decreases exponentially as r increases (see the end line from above in Figure

5.7(b)).


H12 = 0.0031 H13 = 0.0015 H14 = 0.0010 H15 = 0.0007

H22 = 0.0035 H23 = 0.0015 H24 = 0.0009 H25 = 0.0006

H32 = 0.0033 H33 = 0.0016 H34 = 0.0010 H35 = 0.0006

H42 = 0.0029 H43 = 0.0015 H44 = 0.0009 H45 = 0.0006

H52 = 0.0024 H53 = 0.0013 H54 = 0.0008 H55 = 0.0006

H62 = 0.0018 H63 = 0.0010 H64 = 0.0006 H65 = 0.0005

H72 = 0.0012 H73 = 0.0007 H74 = 0.0004 H75 = 0.0003

H82 = 0.0006 H83 = 0.0003 H84 = 0.0002 H85 = 0.0002

H16 = 0.0006 H17 = 0.0005 H18 = 0.0004 H19 = 0.0004

H26 = 0.0005 H27 = 0.0004 H28 = 0.0004 H29 = 0.0005

H36 = 0.0005 H37 = 0.0004 H38 = 0.0004 H39 = 0.0004

H46 = 0.0005 H47 = 0.0004 H48 = 0.0004 H49 = 0.0004

H56 = 0.0004 H57 = 0.0004 H58 = 0.0003 H59 = 0.0004

H66 = 0.0004 H67 = 0.0003 H68 = 0.0003 H69 = 0.0003

H76 = 0.0002 H77 = 0.0002 H78 = 0.0002 H79 = 0.0002

H86 = 0.0001 H87 = 0.0001 H88 = 0.0001 H89 = 0.0001.

สำนกหอ


109



สำนกหอ


110



(a)

(b)





สำนกหอ


111


is similar to that when b = 0.01, b = 0.05 and b = 0.1 m−1 but it decreases rapidly

when (r, z) = (30, 20) and decreases slowly as r increases and it decreases rapidly

and increases slowly as r increases when z = 40 m (see the second and third line

from above in Figure 5.9(b)). Moreover, the value of magnetic field for the case

b = 0.2 m−1 is greater than the case when b = 0.05 m−1 and b = 0.1 m−1. For

z = 0, 60, 80, 100, 120, 140 and 160 m, the value of magnetic field decreases expo-

nentially as r increases.


H12 = 0.0030 H13 = 0.0015 H14 = 0.0010 H15 = 0.0007

H22 = 0.0035 H23 = 0.0016 H24 = 0.0009 H25 = 0.0006

H32 = 0.0034 H33 = 0.0017 H34 = 0.0010 H35 = 0.0007

H42 = 0.0029 H43 = 0.0016 H44 = 0.0010 H45 = 0.0007

H52 = 0.0024 H53 = 0.0013 H54 = 0.0008 H55 = 0.0006

H62 = 0.0018 H63 = 0.0010 H64 = 0.0007 H65 = 0.0005

H72 = 0.0012 H73 = 0.0007 H74 = 0.0005 H75 = 0.0003

H82 = 0.0006 H83 = 0.0003 H84 = 0.0002 H85 = 0.0002

H16 = 0.0006 H17 = 0.0005 H18 = 0.0004 H19 = 0.0004

H26 = 0.0005 H27 = 0.0004 H28 = 0.0004 H29 = 0.0005

H36 = 0.0005 H37 = 0.0004 H38 = 0.0004 H39 = 0.0005

H46 = 0.0005 H47 = 0.0004 H48 = 0.0004 H49 = 0.0004

H56 = 0.0005 H57 = 0.0004 H58 = 0.0004 H59 = 0.0004

H66 = 0.0004 H67 = 0.0003 H68 = 0.0003 H69 = 0.0003

H76 = 0.0003 H77 = 0.0002 H78 = 0.0002 H79 = 0.0002

H86 = 0.0001 H87 = 0.0001 H88 = 0.0001 H89 = 0.0001.

สำนกหอ


112



สำนกหอ


113



(a)

(b)





สำนกหอ


114


is similar to that when b = 0.01, 0.05, 0.1 and b = 0.2 m−1 but it decreases rapidly

when (r, z) = (30, 20) and decreases slowly as r increases and it decreases rapidly

and increases slowly as r increases when z = 40 m and z = 60 m (see the second,

third and fourth line from above in Figure 5.11(b)). Moreover, the value of mag-

netic field for the case b = 0.3 m−1 is greater than the case when b = 0.05, 0.1 and

b = 0.2 m−1. For z = 0, 80, 100, 120, 140 and 160 m, the value of magnetic field

decreases exponentially as r increases.

สำนกหอ


115



(a) (b)

(c) (d)

(e)



m−1 (d) b = 0.2 m−1 (e) b = 0.3 m−1

สำนกหอ


116

From Figure 5.12(a) to (e), when b = 0.01, 0.05, 0.1, 0.2 and 0.3 m−1, re-

spectively, we can see that the value of magnetic field decreases exponentially as

r increases and it decreases as z increases as well. The value of magnetic field

is highest when b = 0.01 m−1 and it decreases when b = 0.05 m−1 and b = 0.1

m−1, respectively and increases slowly when b = 0.2 m−1 and b = 0.3 m−1, respec-

tively. These results are caused by the conductive of the ground and the vertically

location of its.

สำนกหอ


117



bottom).

(a) (b)

(c) (d)

(e)



(d) b = 0.2 m−1 (e) b = 0.3 m−1

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118


creases exponentially as r increases. However, when b = 0.1 m−1 and b = 0.2 m−1,

it starts to decrease rapidly and increases slowly when z = 40 m as we can see in

Figure 5.13(c) and 5.13(d). It then also has the same manner when z = 40 m and

z = 60 m as in Figure 5.13(e).

สำนกหอ


119

(a) (b)

(c) (d)

(e)


is fixed and (a) b = 0.01 m−1 (b) b = 0.05 m−1 (c) b = 0.1 m−1 (d) b = 0.2 m−1

(e) b = 0.3 m−1

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120

Figure 5.14, represents the value of magnetic field when r is fixed (10, 30, . . . ,

and 170) and b = 0.01, 0.05, 0.1, 0.2 and 0.3 m−1, respectively, we can see that the

value of magnetic field decreases from b = 0.01 to b = 0.1 m−1 and it increases

slowly from b = 0.1 to b = 0.3 m−1. The top line represents the value of magnetic

field when r = 10 m and the bottom line represents the value of magnetic field

when r = 170 m. Figure 5.14(c),(d) and (e) show a different behavior of magnetic

field when b = 0.1, 0.2 and 0.3 m−1, respectively. The value of magnetic field drops

down when z = 20 m then increases slowly and drops again as z increases.

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121

(a) (b)

(c) (d)


source when b varies from 0.01, 0.05, 0.1, 0.2 and 0.3 m−1 and z is fixed. (a) z=20

m (b) z=60 m (c) z=100 m (d) z=140 m

Figure 5.15(a) to (d), represents the value of magnetic field which are plot-

ted against r when b varies and z is fixed at 20, 60, 100 and 140m, respectively. We

can see that the value of magnetic field where b = 0.01, 0.05, 0.1, 0.2 and 0.3 m−1

decreases exponentially as r increases and it has similar values when z increases

because the value of magnetic field decreases to zero and has value near zero when

z increases as b varies. From Figure 5.15(b), (c) and (d), the three curves repre-

senting the value of magnetic field when b = 0.1, 0.2 and 0.3 m−1have a different

behavior from the others. The value of magnetic field is greater than that when

b = 0.05 m−1, as we can see the curves cross those two lines for b = 0.05 m−1.

These results take place according to the conductive of ground and the spacing

distance between source and receiver.

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122



(a) (b)

(c) (d)

(e)



= 0.1 m−1 (d) b = 0.2 m−1 (e) b = 0.3 m−1

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123

From Figure 5.16(a) to (e), when b = 0.01, 0.05, 0.1, 0.2 and 0.3 m−1, the

red color shows the area when the value of magnetic field is high and the blue color

shows the area when the value of magnetic field is low. The value of magnetic

field decreases from b = 0.01 to b = 0.1 m−1 and it increases slowly from b = 0.1

to b = 0.3 m−1, as we can see in Figure 5.16(a) to (e).

5.2 Case of an Exponentially Increasing Conductivity

Turning to the case of increasing conductivity σ = σ0+(σ1−σ0)e−bz when

σ0 > σ1, σ1 = 0.5 and σ0 = 1.5 (S/m), the numerical solutions of the magnetic

field at each node is calculated by using MATLAB program.


H12 = 0.0056 H13 = 0.0035 H14 = 0.0025 H15 = 0.0020

H22 = 0.0050 H23 = 0.0033 H24 = 0.0025 H25 = 0.0020

H32 = 0.0043 H33 = 0.0029 H34 = 0.0022 H35 = 0.0018

H42 = 0.0035 H43 = 0.0024 H44 = 0.0019 H45 = 0.0015

H52 = 0.0028 H53 = 0.0019 H54 = 0.0015 H55 = 0.0012

H62 = 0.0021 H63 = 0.0014 H64 = 0.0011 H65 = 0.0009

H72 = 0.0014 H73 = 0.0009 H74 = 0.0007 H75 = 0.0006

H82 = 0.0007 H83 = 0.0005 H84 = 0.0004 H85 = 0.0003

H16 = 0.0016 H17 = 0.0013 H18 = 0.0011 H19 = 0.0009

H26 = 0.0016 H27 = 0.0013 H28 = 0.0011 H29 = 0.0009

H36 = 0.0015 H37 = 0.0012 H38 = 0.0010 H39 = 0.0008

H46 = 0.0013 H47 = 0.0010 H48 = 0.0008 H49 = 0.0006

H56 = 0.0010 H57 = 0.0008 H58 = 0.0007 H59 = 0.0005

H66 = 0.0008 H67 = 0.0006 H68 = 0.0005 H69 = 0.0004

H76 = 0.0005 H77 = 0.0004 H78 = 0.0003 H79 = 0.0003

H86 = 0.0003 H87 = 0.0002 H88 = 0.0002 H89 = 0.0001.

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124



สำนกหอ


125



(a)

(b)





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126






H12 = 0.0078 H13 = 0.0056 H14 = 0.0045 H15 = 0.0037

H22 = 0.0067 H23 = 0.0055 H24 = 0.0047 H25 = 0.0040

H32 = 0.0052 H33 = 0.0043 H34 = 0.0038 H35 = 0.0033

H42 = 0.0040 H43 = 0.0032 H44 = 0.0028 H45 = 0.0025

H52 = 0.0030 H53 = 0.0023 H54 = 0.0020 H55 = 0.0018

H62 = 0.0022 H63 = 0.0016 H64 = 0.0014 H65 = 0.0012

H72 = 0.0014 H73 = 0.0010 H74 = 0.0009 H75 = 0.0008

H82 = 0.0007 H83 = 0.0005 H84 = 0.0004 H85 = 0.0004

H16 = 0.0031 H17 = 0.0025 H18 = 0.0020 H19 = 0.0015

H26 = 0.0034 H27 = 0.0028 H28 = 0.0022 H29 = 0.0014

H36 = 0.0028 H37 = 0.0023 H38 = 0.0018 H39 = 0.0012

H46 = 0.0021 H47 = 0.0018 H48 = 0.0013 H49 = 0.0009

H56 = 0.0016 H57 = 0.0013 H58 = 0.0010 H59 = 0.0007

H66 = 0.0011 H67 = 0.0009 H68 = 0.0007 H69 = 0.0005

H76 = 0.0007 H77 = 0.0006 H78 = 0.0004 H79 = 0.0003

H86 = 0.0003 H87 = 0.0003 H88 = 0.0002 H89 = 0.0002.

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127



สำนกหอ


128



(a)

(b)





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129


similar to that when b = 0.01 m−1. However, the value of magnetic field is greater

than the case as b = 0.01 m−1. From Figure 5.20(b), when b = 0.05 m−1, the

value of magnetic field decreases exponentially when r increases and it is greater

than the case as b = 0.01 m−1.


H12 = 0.0095 H13 = 0.0075 H14 = 0.0063 H15 = 0.0054

H22 = 0.0076 H23 = 0.0068 H24 = 0.0061 H25 = 0.0054

H32 = 0.0055 H33 = 0.0048 H34 = 0.0045 H35 = 0.0040

H42 = 0.0041 H43 = 0.0035 H44 = 0.0032 H45 = 0.0029

H52 = 0.0031 H53 = 0.0025 H54 = 0.0022 H55 = 0.0020

H62 = 0.0022 H63 = 0.0017 H64 = 0.0015 H65 = 0.0014

H72 = 0.0015 H73 = 0.0011 H74 = 0.0010 H75 = 0.0009

H82 = 0.0007 H83 = 0.0005 H84 = 0.0005 H85 = 0.0004

H16 = 0.0046 H17 = 0.0038 H18 = 0.0029 H19 = 0.0020

H26 = 0.0046 H27 = 0.0038 H28 = 0.0029 H29 = 0.0018

H36 = 0.0035 H37 = 0.0029 H38 = 0.0021 H39 = 0.0013

H46 = 0.0025 H47 = 0.0021 H48 = 0.0016 H49 = 0.0010

H56 = 0.0018 H57 = 0.0015 H58 = 0.0011 H59 = 0.0007

H66 = 0.0012 H67 = 0.0010 H68 = 0.0008 H69 = 0.0005

H76 = 0.0008 H77 = 0.0006 H78 = 0.0005 H79 = 0.0003

H86 = 0.0004 H87 = 0.0003 H88 = 0.0002 H89 = 0.0002.

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130



สำนกหอ


131



(a)

(b)





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132

From Figure 5.22(a), when b = 0.1 m−1, the behavior of magnetic field

is similar to that when b = 0.01 m−1 and b = 0.05 m−1. However, the value of

magnetic field is greater than the case as b = 0.01 m−1 and b = 0.05 m−1. From

Figure 5.22(b), when b = 0.1 m−1, the value of magnetic field decreases exponen-

tially when r increases and it is greater than the case as b = 0.01 m−1 and b = 0.05

m−1.


H12 = 0.0107 H13 = 0.0090 H14 = 0.0079 H15 = 0.0069

H22 = 0.0077 H23 = 0.0073 H24 = 0.0067 H25 = 0.0060

H32 = 0.0055 H33 = 0.0050 H34 = 0.0047 H35 = 0.0043

H42 = 0.0042 H43 = 0.0036 H44 = 0.0033 H45 = 0.0031

H52 = 0.0032 H53 = 0.0026 H54 = 0.0023 H55 = 0.0021

H62 = 0.0023 H63 = 0.0018 H64 = 0.0016 H65 = 0.0015

H72 = 0.0015 H73 = 0.0011 H74 = 0.0010 H75 = 0.0009

H82 = 0.0007 H83 = 0.0005 H84 = 0.0005 H85 = 0.0004

H16 = 0.0059 H17 = 0.0048 H18 = 0.0037 H19 = 0.0024

H26 = 0.0052 H27 = 0.0043 H28 = 0.0032 H29 = 0.0020

H36 = 0.0038 H37 = 0.0031 H38 = 0.0023 H39 = 0.0014

H46 = 0.0027 H47 = 0.0022 H48 = 0.0017 H49 = 0.0011

H56 = 0.0019 H57 = 0.0016 H58 = 0.0012 H59 = 0.0008

H66 = 0.0013 H67 = 0.0011 H68 = 0.0008 H69 = 0.0006

H76 = 0.0008 H77 = 0.0007 H78 = 0.0005 H79 = 0.0004

H86 = 0.0004 H87 = 0.0003 H88 = 0.0002 H89 = 0.0002.

สำนกหอ


133



สำนกหอ


134



(a)

(b)





สำนกหอ


135


similar to that when b = 0.01, 0.05 and 0.1 m−1. However, the value of magnetic

field is greater than the case as b = 0.01, 0.05 and 0.1 m−1. From Figure 5.24(b),

when b = 0.2 m−1, the value of magnetic field decreases exponentially when r

increases and it is greater than the case as b = 0.01, 0.05 and 0.1 m−1.


H12 = 0.0110 H13 = 0.0093 H14 = 0.0082 H15 = 0.0072

H22 = 0.0076 H23 = 0.0071 H24 = 0.0066 H25 = 0.0059

H32 = 0.0055 H33 = 0.0049 H34 = 0.0046 H35 = 0.0043

H42 = 0.0042 H43 = 0.0035 H44 = 0.0033 H45 = 0.0030

H52 = 0.0031 H53 = 0.0025 H54 = 0.0023 H55 = 0.0021

H62 = 0.0023 H63 = 0.0018 H64 = 0.0016 H65 = 0.0014

H72 = 0.0015 H73 = 0.0011 H74 = 0.0010 H75 = 0.0009

H82 = 0.0007 H83 = 0.0005 H84 = 0.0005 H85 = 0.0004

H16 = 0.0062 H17 = 0.0051 H18 = 0.0039 H19 = 0.0025

H26 = 0.0051 H27 = 0.0042 H28 = 0.0032 H29 = 0.0019

H36 = 0.0037 H37 = 0.0031 H38 = 0.0023 H39 = 0.0014

H46 = 0.0027 H47 = 0.0022 H48 = 0.0016 H49 = 0.0010

H56 = 0.0019 H57 = 0.0015 H58 = 0.0012 H59 = 0.0008

H66 = 0.0013 H67 = 0.0011 H68 = 0.0008 H69 = 0.0005

H76 = 0.0008 H77 = 0.0007 H78 = 0.0005 H79 = 0.0004

H86 = 0.0004 H87 = 0.0003 H88 = 0.0002 H89 = 0.0002.

สำนกหอ


136



สำนกหอ


137



(a)

(b)





สำนกหอ


138


similar to that when b = 0.01, 0.05, 0.1 and b = 0.2 m−1 but it is smaller than

the case as b = 0.2 m−1. From Figure 5.26(b), when b = 0.3 m−1, the value of

magnetic field decreases exponentially as r increases and it is smaller than the

case as b = 0.2 m−1.

สำนกหอ


139



(a) (b)

(c) (d)

(e)



m−1 (d) b = 0.2 m−1 (e) b = 0.3 m−1

สำนกหอ


140

From Figure 5.27(a) to (e), when b = 0.01, 0.05, 0.1, 0.2 and 0.3 m−1, re-

spectively, we can see that the value of magnetic field decreases exponentially as

r increases and it decreases as z increases as well. The value of magnetic field

increases when b increases and it decreases slowly from b = 0.2 m−1 to b = 0.3

m−1. This results is caused by the conductive of the ground and the vertically

location of its which is as same as the case of decreasing conductivity profile.

สำนกหอ


141



bottom).

(a) (b)

(c) (d)

(e)



(d) b = 0.2 m−1 (e) b = 0.3 m−1

สำนกหอ


142


creases exponentially as r increases. The value of magnetic field increases when

b increases and it decreases slowly from b = 0.2 to b = 0.3 m−1 (we can see in

Figure 5.28(d) and (e)).

สำนกหอ


143

(a) (b)

(c) (d)

(e)


is fixed and (a) b = 0.01 m−1 (b) b = 0.05 m−1 (c) b = 0.1 m−1 (d) b = 0.2 m−1

(e) b = 0.3 m−1

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144

Figure 5.29, represents the value of magnetic field when r is fixed (10, 30, . . . ,

and 170) and b = 0.01, 0.05, 0.1, 0.2 and 0.3 m−1, respectively, we can see that the

value of magnetic field decreases as r increases. The value of magnetic field in-

creases when b increases and it decreases slowly from b = 0.2 to b = 0.3 m−1 (we

can see in Figure 5.29(d) and (e)).

สำนกหอ


145

(a) (b)

(c) (d)


source when b varies from 0.01, 0.05, 0.1, 0.2 and 0.3 m−1 and z is fixed. (a) z=20

m (b) z=60 m (c) z=100 m (d) z=140 m

Figure 5.30(a) to (d), represents the values of magnetic field which are

plotted against r when b varies and z is fixed at 20, 60, 100 and 140 m, respectively.

We can see that the value of magnetic field where b = 0.01, 0.05, 0.1, 0.2 and

0.3 m−1 decreases exponentially as r increases and it has similar values when z

increases because the value of magnetic field decreases to zero and has value near

zero when z increases as b varies. From Figure 5.30(b), (c) and (d), the curve

representing the values of magnetic field which are plotted against r when b = 0.3

m−1 have a different behavior from the others. The value of magnetic field is

smaller than that when b = 0.2 m−1, as we can see the curve cross the line for

b = 0.2 m−1. These results occur according to the conductive of ground and the

spacing distance between source and receiver.

สำนกหอ


146



(a) (b)

(c) (d)

(e)



= 0.1 m−1 (d) b = 0.2 m−1 (e) b = 0.3 m−1

สำนกหอ


147

From Figure 5.31(a) to (e), when b = 0.01, 0.05, 0.1, 0.2 and 0.3 m−1, the

red color shows the area when the value of magnetic field is high and the blue color

shows the area when the value of magnetic field is low. The value of magnetic field

increases when b increases and it decreases slowly from b = 0.2 m−1 to b = 0.3

m−1, as we can see in Figure 5.31(a) to (e).

Summarize

In this chapter, finite element method is used to approximate the solution

of partial differential equation. The Maxwell’s equation is our governing equation

that can be used to find magnetic field. Under the boundary conditions and the

conductivity of the ground as σ(z) = σ0 + (σ1 − σ0)e−bz we obtain the behavior of

magnetic field decreases to zero when the depth of soil increases. As well as the

case of increasing the space between source-receiver, the magnetic field decreases

to zero too. The value of b is an important role for the conduction of ground and

effect to the magnetic field quantities as well. The comparision of the quantities

of magnetic field for the case of σ as an increasing function is higher than the case

of σ as a decreasing function according to the advantage of the DC source that

better reflex at very high depth than on the ground surface.

สำนกหอ


Chapter 6

Results and Discussion

A mathematical model of the magnetic fields is conducted by using par-

tial differential equations. Numerical solutions of partial differential equations are

computed by using numerical techniques of the finite element method (FEM) to

find the value of the magnetic field at various locations of the ground. In our

models, the ground have exponentially decreasing and increasing conductivities

profiles. The Galerkin’s Method of Weighted Residuals is applied to finite element

method. There is a source providing a DC voltage and a receiver on the ground

surface which picks up the signal from r = 10 m to r = 190 m. We begin our study

with the relationship between magnetic and electric fields. The vector equations

for computing magnetic and electric field is very difficult. Thus, in most research,

we change the vector equations into scalar equations by considering the compo-

nents of the vector that we inject to the ground. We finally arrive with partial

differential equation describing the system and it is surprisingly that our equation

is independent from the electric current. The numerical results is calculated and

plotted by using MATLAB R2009a program to show the behavior of magnetic

field at different depths and distances from source point. In our research, the

behavior of magnetic field decreases to zero when the depth of soil increases. As

well as the case of increasing the space between source-receiver, the magnetic field

decreases to zero too. The value of b is an important role for the conduction of

ground and effect to the magnetic field quantities as well. The comparision of

the quantities of magnetic field for the case of σ = σ0e−bz is not much different

to the case of σ = σ0 + (σ1 − σ0)e−bz for a decreasing function. But the rate of

decrease of the magnetic field for the case of σ = σ0e−bz is greater than the case of

σ = σ0+(σ1−σ0)e−bz for a decreasing function because the conduction of ground

148

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149

in the case of σ = σ0 + (σ1 − σ0)e−bz is higher than the case of σ = σ0e

−bz. The

comparision of the quantities of magnetic field for the case of σ = σ0e−bz is lower

than the case of σ = σ0 + (σ1 − σ0)e−bz for an increasing function. The rate of

increase of the magnetic field for the case of σ = σ0e−bz is lower than the case of

σ = σ0+(σ1−σ0)e−bz for an increasing function because the conduction of ground

in the case of σ = σ0 + (σ1 − σ0)e−bz better than the case of σ = σ0e

−bz.

6.1 Future Works

Even though the work presented in this thesis provides interesting ideas

about the solution to the forward problems in the cause the magnetic field re-

sponse, the issues that we dealt with suggest numerous avenues for possible exten-

sions and future works. In the area of magnetometric resistivity methods described

in this thesis, the following outline is a list of interesting future directions that

require further investigation:

• On the magnetometric resistivity method, numerical solution of the magnetic

field response from a multilayered earth containing buried electrodes can be

derived by using the finite element method of the Galerkin’s Method of

Weighted Residuals described in Chapters 3, 4 and 5.

• On the magnetometric resistivity method, numerical solution of the magnetic

field response from a multilayered earth containing buried electrodes can be

derived by using the finite element method that make a triangulation of

domain.

• The numerical solution of the magnetic field response from a multilayered

earth containing buried electrodes having conductivity is functions of radial

r and depth z can be derived by using the finite element method of the

Galerkin’s Method of Weighted Residuals.

สำนกหอ


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[36] J. Velimsky and Z. Martinec, Time-domain,spherical harmonic-finite element

approach to transient three-dimensional geomagnetic induction in a spherical

heterogeneous Earth, Geophysics, 161(1)(2005), 81− 101.

[37] S. Yooyuanyong and N. Chumchob, Mathematical modeling of electromag-

netic sounding for a conductive 3-D circular body embedded in half-space,

Proceedings of the Third Asian Mathematical Conference, (2000), 590− 603.

[38] Wikipedia, Stokes stream function, Search 24/06/2014, http : //en.wikipedia

.org/wiki/Stokes stream function.

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Appendix

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60 80 100 120 140 0 20 40 60 80 100 120 140 160 180 200 220 240 260 280 300320 340 360 380 ซโอดละลาย(มก./ล.)ระยะเวลา(วน) a b c d e f g B A 73 กข

0 0 0 0 0 0 0 0 20 h 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0A5 1 6 0 0 0 0 0 0 0 0 0 0

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27 s1 1 6 0 20 h 0 0 0 0 0 0 0 0 0 0 0 0 340 h 0 s2 1 6 340 h 0 0 0 0 0 0 0 0 0 0 0 00 0 0 s3 1 6 0 0 0 0 0 0 0 0 0 0 0 0 0 0 340 h 0 s4 1 6 0 20 h 0 0 0 0 0 0 0 0 0 0 0 00 0 t 1 0 0 1 Since we know the value of the magnetic field at the boundary H1

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ผลการตรวจสอบ abstract สาหรบงานประชม AMM 2015

จาก: [email protected] ในนามของ [email protected]สงเมอ: 11 มนาคม 2558 20:38:02ถง: [email protected]

เรยน คณปรยานช ตนนรกษ ในนามของคณะกรรมการจดการประชมวชาการคณตศาสตรประจาป 2558 ครงท 20 ขอแจงใหทานทราบวากรรมการฝายวชาการของการประชม ฯ ไดพจารณาบทคดยอของทานในหวขอ Finite Element Magnetic Field Response of an Exponential Conductivity GroundProfile เรยบรอยแลว และมความยนดขอเชญทานเขารวมประชมและเสนอผลงานในหวขอดงกลาวในการประชมวชาการคณตศาสตรประจาป 2558 ครงท20 ทงน ไดกาหนดเวลาสาหรบการนาเสนอผลงาน 15 นาท สวนกาหนดการนาเสนอจะแจงใหทานทราบตอไป นอกจากน ทานสามารถรบทราบขอมลตาง ๆ ของการประชมครงนไดทางเวบไซต http://www.amm2015.com/ อนง หากทานมขอสงสยประการใด กรณาตดตอคณะกรรมการจดการประชม ฯไดท อ.ดร. พรทรพย พรสวสด โทรศพท 084-5633225 ในเวลาราชการ หรออเมลมาท [email protected] ขอแสดงความนบถอ กรรมการวชาการฯ

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Dr. Emil Minchev Managing Editor, President of Hikari Ltd http://www.m-hikari.com Dear Professor Yooyuanyong, I am happy to inform you that after a positive referee report your paper: "Finite Element Magnetic Field Response of an Exponential Conductivity Ground Profile" (with Priyanuch Tunnurak, Nairat Kanyamee) has been accepted for publication in Applied Mathematical Sciences. Please SEND me by e-mail the final Latex file and PDF of your paper as soon as possible. Reprints will be provided to you after the publication of your paper. Yours sincerely, Emil Minchev President of Hikari Ltd Managing Editor of Applied Mathematical Sciences

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Applied Mathematical Sciences, Vol. 9, 2015, no. 52, 2579 - 2594HIKARI Ltd, www.m-hikari.com

http://dx.doi.org/10.12988/ams.2015.52116

Finite Element Magnetic Field Response of

an Exponential Conductivity Ground Profile

Priyanuch Tunnurak, Nairat Kanyamee and Suabsagun Yooyuanyong

Department of Mathematics, Faculty of ScienceSilpakorn University and Centre of Excellence in Mathematics, Thailand

Copyright c© 2015 Priyanuch Tunnurak, Nairat Kanyamee and Suabsagun Yooyuanyong.

This is an open access article distributed under the Creative Commons Attribution License,

which permits unrestricted use, distribution, and reproduction in any medium, provided the

original work is properly cited.

Abstract

In this paper, mathematical model of finite element method for themagnetic field of an exponential conductivity ground profile is presentedand computed to find the magnetic field at various locations by assumingthat the Earth structure having exponential conductivity profile. Thereis a source providing a DC voltage and a receiver on the ground surface.Numerical technique the finite element method (FEM) is introducedby using the Galerkin’s method of Weighted Residuals to find approx-imate solutions of partial differential equation. Matlab programing isconducted to calculate and plot graphs of magnetic field at various lo-cations. The results perform very well to the intensity of magnetic fieldfor cross sections of ground structure.

Mathematics Subject Classification: 86A25

Keywords: Finite element, Magnetic, Galerkin

1 Introduction

Currently, human studied the Earth structure widely in order to utilizethe natural resources embedded beneath the Earth for developing the agricul-tural sector and industrial sectors in their countries. They use knowledge ofgeophysics which is a branch of science concerned with the Earth survey. The

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2580 Priyanuch Tunnurak et al.

survey uses mathematics, physics and the physical properties of the Earth suchas the resistivity, conductivity, electric potential, magnetic field and electricfield to search for the natural resources.

We create a mathematical model by using magnetometric resistivity methodto find the value of magnetic field beneath the Earth surface. In 2003, Chenand Oldenburg[7] assume that the Earth structure consists of horizontallystratified layers having constant conductivity at certain depths except thelast layer where the conductivity having the same varying through the restof the layer. They derived the magnetic field directly by solving a bound-ary value problem of a horizontally stratified layered Earth with homogeneouslayers. However, in the real situation there are cases where the subsurfaceconductivities vary exponentially, linearly or binomially with depth. Thereexists a considerable amount of research about mathematical modeling whichassumes that the Earth structure consists of horizontally stratified multilayerwith one or more layers having exponentially, linearly or binomially varyingconductivities at certain depths except the last layer where the conductivityhaving the same varying through the rest of the layer. Stoyer and Wait[28]studied the problem of computing apparent resistivity for a structure with ahomogeneous overburden overlying a medium whose resistivity varies exponen-tially with depth. Banerjee et al.[1] gave expressions for apparent resistivityof a multilayered Earth with a layer having exponentially varying conductiv-ity. Kim and Lee[14] derived a new resistivity kernel function for calculatingapparent resistivity of a multilayered Earth with layers having exponentiallyvarying conductivities. Siew and Yooyuanyong[29] studied the electromagneticresponse of a thin disk beneath an inhomogeneous conductive overburden andexpressions for the electric fields in the overburden. Ketchanwit[15] studied theEarth surface layers using time-domain electromagnetic field by constructingthree mathematical models having exponentially varying and constant vary-ing conductivities. Sripunya[30] derived solutions of the steady state magneticfield due to a DC current source in a layered Earth with some layer havingexponentially or binomially or linearly varying conductivity.

In this paper, mathematical model is presented by using numerical tech-niques for finding approximate solutions. The finite element method (FEM)is used to find the numerical solutions of the magnetic field under the Earthsurface. We assume that the Earth structure contains only one layer havingexponential conductivity (σ(z) = σ0+(σ1−σ0)e

−bz). This method is differentfrom the Hankel transform approach which is difficult to solve for some complexproblems such as all the research mentioned above. There are a few researchusing FEM by applying the Galerkin’s method of Weighted Residuals to findthe solution of the magnetic field. For instance, Lee[16] presented a numericalmethod of computing the electromagnetic response of two-dimensional Earthmodels to an oscillating magnetic dipole. Velimsky and Martince[36] intro-

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Finite element magnetic field response 2581

duced a time-domain method to solve the problem of geomagnetic inductionin a heterogeneous Earth excited by variations of the ionospheric and mag-netospheric currents with arbitrary spatiotemporal characteristics. Mitsuhataand Uchida[20] presented a finite element algorithm for computing magneticfield response for 3D conductivity structures. Therefore we are interested ap-proximate techniques in finding the magnetic field beneath the Earth by usingthe Galerkin’s method of Weighted Residuals.

2 Mathematical Formulations

In this section, we use finite element method (FEM) for constructing ap-proximate solutions of our problems. Assuming that the Earth structure con-tains only one layer having exponential conductivity (σ(z) = σ0+(σ1−σ0)e

−bz)and there are a source providing a DC voltage and a receiver on the groundsurface which picks up the signal from r = 10 m to r = 190 m as shown inFigure 2.1.

Figure 2.1: Geometric model of the Earth structure.

We define z as the depth of an object from the Earth surface (meter), r asthe distance between source and receiver of magnetic field on the Earth surface(meter) and σ(z) as the conductivity of the medium which is a function of z(S/m), where σ0, σ1 and b are positive constants.

From Maxwell’s equations , the relationship between the electric and mag-netic fields[29,30,31,37] written in cylindrical coordinates (r, φ, z) is as follows.

∇× �E = �0 (2.1)

and∇× �H = σ �E, (2.2)

where �E is the electric field vector, �H is the magnetic field vector, σ is the

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conductivity of the medium and ∇ is the gradient operator in cylindrical co-ordinates (r, φ, z) [17,27] defined by

∇ =∂

∂rer +

1

r

∂

∂φeφ +

∂

∂zez,

where er is the unit vector in radial direction (r), eφ is the unit vector in thedirection of φ, ez is the unit vector in the direction of z.

Substituting equation (2.2) into (2.1) , we obtain

∇× 1

σ(∇× �H) = �0. (2.3)

Substitute equation ∇× 1σ(∇× �H) in cylindrical coordinates (r, φ, z) [17,27]

into (2.3) , we obtain

1

r

[∂

∂φ

(1

r

(1

σ

∂(rHφ)

∂r− 1

σ

∂Hr

∂φ

))− ∂

∂z

(1

σ

∂Hr

∂z− 1

σ

∂Hz

∂r

)]er

+

[∂

∂z

(1

r

(1

σ

∂Hz

∂φ− 1

σ

∂(rHφ)

∂z

))− ∂

∂r

(1

r

(1

σ

∂(rHφ)

∂r− 1

σ

∂Hr

∂φ

))]eφ

+1

r

[∂

∂r

(1

σ

∂Hr

∂z− 1

σ

∂Hz

∂r

)− ∂

∂φ

(1

r

(1

σ

∂Hz

∂φ− 1

σ

∂(rHφ)

∂z

))]ez = �0, (2.4)

where Hr, Hφ and Hz are the components of �H in er, eφ and ez directions,respectively. Since the magnetic field is axisymmetric, it depends only on rand z and not on the azimuth φ and from electromagnetic theory, we knowthe magnetic field has only the azimuthal component, i.e. �H = Hφ(r, z)eφ[19].Simplifying equation (2.4) yields

1

σ

∂2H

∂z2+

∂H

∂z

∂

∂z

(1

σ

)+

1

σ

[1

r

∂2(rH)

∂r2+

∂

∂r

(1

r

)∂(rH)

∂r

]= 0.

We denote conductivity σ as a function of depth z only, i.e. σ = σ(z), andwe now have

∂2H

∂z2+ σ

∂H

∂z

∂

∂z

(1

σ

)+

∂2H

∂r2+

1

r

∂H

∂r− 1

r2H = 0. (2.5)

Since the Laplace equation and the problem is axisymmetric, our problembecomes

ΔH + σ∂H

∂z

∂

∂z

(1

σ

)− 1

r2H = 0. (2.6)

The next step, we use finite element method to establish a numerical solu-tion of our problem. We apply the Galerkin’s Method of Weighted Residuals

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to equation (2.6).We transform equation (2.6) into weak formulation to find H ∈ H1. Let

V = { v ∈ H1 : v is a continuous function on Ω, ∂v∂r

and ∂v∂z

are piecewisecontinuous on Ω and v = 0 on ∂Ω }.The weak formulation of equation (2.6) is denoted by

(ΔH, v) +

(σ∂H

∂z

∂

∂z

( 1

σ

), v

)− (

1

r2H, v) = 0 , v ∈ V

or ∫Ω

ΔHvdΩ +

∫Ω

σ∂H

∂z

∂

∂z

( 1

σ

)vdΩ−

∫Ω

1

r2HvdΩ = 0. (2.7)

By Green’s identity[32] and v = 0 on ∂Ω, we obtain

−∫Ω

∇H · ∇vdΩ +

∫Ω

σ∂H

∂z

∂

∂z

( 1

σ

)vdΩ−

∫Ω

1

r2HvdΩ = 0. (2.8)

Using cylindrical co-ordinates (r, φ, z) [21] and since the problem is ax-isymmetric and H has only the azimuthal component in cylindrical coordinate,equation (2.8) becomes

−∫Ω

r∇H · ∇vdrdz +

∫Ω

rσ∂H

∂z

∂

∂z

( 1

σ

)vdrdz −

∫Ω

1

rHvdrdz = 0, (2.9)

where Ω is the 2D cross-section of domain Ω (φ is fixed), i.e. Ω = {(r, z), 10 ≤r ≤ 190, 0 ≤ z ≤ 180}.

Next we consider the two dimensional domain of equation (2.9). By divid-ing the domain into rectangular elements, we discretize r into 9 subintervalsequally, discretize z into 9 subintervals equally and (ri, zj) is a node of Ω onthe non overlapping rectangles such that the horizontal and vertical edges ofthese rectangles are parallel to the r and z coordinate axes,respectively, i.e.

ri = 10 + 20i , i = 0, . . . , 9,

zj = 20j , j = 0, . . . , 9.

Since the form of equation (2.9) suggests that the finite elements can havean arbitrary shape and position in space computing integrals over their ele-ment domains is a bit tricky. To overcome this difficulty, one uses a projec-tion method which maps the coordinates of a well known reference elementto the coordinates of an arbitrary element in space by a mapping the valuesrange from -1 to +1, and the reference coordinates are as (ξ1, η1) = (−1,−1),(ξ2, η2) = (1,−1) , (ξ3, η3) = (1, 1) , (ξ4, η4) = (−1, 1) that represent in Figure2.2(b).

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(a) A rectangularelements.

(b) Thereferenceelement.

Figure 2.2: The coordinate transformation (r, z) in terms of the local coordi-nates (ξ, η)

Figure 2.3: The nodes {Hi}100i=1 of the elements.

For simplicity and to avoid any confusion, we use H(Xi), i = 1, 2, . . . , 100for Hi, i = 1, 2, . . . , 100. In other words, we define nodes Xi, i = 1, 2, . . . , 100for (ri, zj), i, j = 0, 1, . . . , 9 as shown in Figure 2.3.For each i = 1, 2, . . . , 100 define ϕj as basis function such that

ϕj(Xi) =

{1 , i = j0 , i �= j

.

A function v ∈ V can be written in the form of linear combination of basisfunction ϕi

v(X) =100∑i=1

αiϕi(X).

We obtain v(Xj) = αj by choosing appropriate values for αj. Equation (2.9)becomes

−(r∇H,∇ϕi) +

(rσ

∂H

∂z

∂

∂z

( 1

σ

), ϕi

)− (

1

rH, ϕi) = 0.

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Substituing (σ(z) = σ0 + (σ1 − σ0)e−bz) where σ0, σ1 and b are positive con-

stants, we obtain

−(r∇H,∇ϕi) +

(br

((σ1 − σ0)e

−bz

σ0 + (σ1 − σ0)e−bz

)∂H

∂z, ϕi

)− (

1

rH, ϕi) = 0,

for i = 1, 2, . . . , 100.Next, we consider the solution in the form of linear combination of basis

function ϕj

H(X) =100∑j=1

Hjϕj(X),

when Hj is the unknown constants to be found.Then equation (2.9) can be written in the form of linear combination as

follows

100∑j=1

Hj

[−∫Ω

r(∇ϕj·∇ϕi)drdz+

∫Ω

br

((σ1 − σ0)e

−bz

σ0 + (σ1 − σ0)e−bz

)∂ϕj

∂zϕidrdz−

∫Ω

1

rϕjϕidrdz

]= 0,

(2.10)for each i = 1, 2, . . . , 100.

We need a transformation from an original element to a reference elementas shown in Figure 2.2

r = rk +h

2(1 + ξ), dr =

h

2dξ,

z = zk +h

2(1 + η), dz =

h

2dη. (2.11)

where k = 0, 1, . . . , 8.The basis functions can be written in the form of ξ and η as follows

N1(ξ, η) =1

4(1− ξ)(1− η),

N2(ξ, η) =1

4(1 + ξ)(1− η),

N3(ξ, η) =1

4(1 + ξ)(1 + η),

N4(ξ, η) =1

4(1− ξ)(1 + η). (2.12)

We transform r, z to ξ, η by using the transformation equation (2.11) to-

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gether with basis functions equation (2.12), so equation (2.10) becomes

100∑j=1

Hj

[−

∫ 1

−1

∫ 1

−1

(rk +

h

2(1 + ξ)

)[ ˆ∂ϕj

∂ξ

ˆ∂ϕi

∂ξ+

ˆ∂ϕj

∂η

ˆ∂ϕi

∂η

]dξdη

+bh

2

∫ 1

−1

∫ 1

−1

(rk +

h

2(1 + ξ)

)( (σ1 − σ0)e−b(zk+

h2(1+η)

)σ0 + (σ1 − σ0)e

−b(zk+

h2(1+η)

))

ˆ∂ϕj

∂ηϕidξdη

− h2

4

∫ 1

−1

∫ 1

−1

ϕjϕi(rk +

h2(1 + ξ)

)dξdη] = 0, (2.13)

for each i = 1, 2, . . . , 100.The value of equation (2.13) can be written in the form of matrix as follows(−A+ B − C

)ui =

(a+ d+ c

),

where (−A + B − C), the stiffness matrix is an 100 × 100 matrix, ui and(a+ d+ c) are 100-vectors for all i = 1, 2, . . . , 100.

3 Numerical Results

Since the Galerkin’s Method of Weighted Residuals was applied to equation(2.5), we obtained the values of magnetic field at various positions of theearth’s structure with one layer having exponentially decreasing conductivityσ = σ0 + (σ1 − σ0)e

−bz when σ0 < σ1. There is a source providing a DCvoltage and a receiver on the ground surface which picks up the signal fromr = 10 m to r = 190 m. We discrete the depth into 9 subintervals equallyof the size h = 20 m, i.e. we consider z = 0, 20, . . . , 180 m. We use constantσ1 = 1.5 S/m, σ0 = 0.5 S/m and b = 0.01, 0.05, 0.1, 0.2 and 0.3 m−1. Thenumerical solutions of the magnetic field at each node is calculated by usingMatlab program.

Graphs of the relationship between magnetic field and distance of receiver fromsource at various depths.

(a) (b) (c)

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(d) (e)

Figure 3.1: The relationship between magnetic field and distance of receiverfrom source when z is fixed for (a) b = 0.01 m−1 (b) b = 0.05 m−1 (c) b =0.1 m−1 (d) b = 0.2 m−1 (e) b = 0.3 m−1

From Figure 3.1(a) to (e), when b = 0.01, 0.05, 0.1, 0.2 and 0.3 m−1, respec-tively, we can see that the value of magnetic field decreases exponentially as rincreases and it decreases as z increases as well. The value of magnetic field ishighest when b = 0.01 m−1 and it decreases when b = 0.05 m−1 and b = 0.1m−1, respectively and increases slowly when b = 0.2 m−1 and b = 0.3 m−1,respectively. These results are caused by the conductive of the ground and thevertically location of its.

(a) (b)

(c) (d)

Figure 3.2: The relationship between magnetic field and distance of receiverfrom source when b varies from 0.01, 0.05, 0.1, 0.2 and 0.3 m−1 and z is fixed.(a) z=20 m (b) z=60 m (c) z=100 m (d) z=140 m

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Figure 3.2(a) to (d) represents the value of magnetic field which are plottedagainst r when b varies and z is fixed at 20, 60, 100 and 140 m, respectively.We can see that the value of magnetic field where b = 0.01, 0.05, 0.1, 0.2 and0.3 m−1 decreases exponentially as r increases and it has similar values whenz increases because the value of magnetic field decreases to zero and has valuenear zero when z increases as b varies. From Figure 3.2(b), (c) and (d), thethree curves representing the value of magnetic field when b = 0.1, 0.2 and 0.3m−1have a different behavior from the others. The value of magnetic field isgreater than that when b = 0.05 m−1, as we can see the curves cross those twolines for b = 0.05 m−1. These results take place according to the conductiveof ground and the spacing distance between source and receiver.

Contour graphs of the relationship between magnetic field and distance ofreceiver from source at various depth are shown as in Figure 3.3.

(a) (b) (c)

(d) (e)

Figure 3.3: Contour graphs of magnetic field at different distances of receiverfrom source and different depths when (a) b = 0.01 m−1 (b) b = 0.05 m−1

(c) b = 0.1 m−1 (d) b = 0.2 m−1 (e) b = 0.3 m−1

From Figure 3.3(a) to (e), when b = 0.01, 0.05, 0.1, 0.2 and 0.3 m−1, thered color shows the area when the value of magnetic field is high and the bluecolor shows the area when the value of magnetic field is low. The value ofmagnetic field decreases from b = 0.01 to b = 0.1 m−1 and it increases slowlyfrom b = 0.1 to b = 0.3 m−1, as we can see in Figure 3.3(a) to (e).

Turning to the case of increasing conductivity σ = σ0+(σ1−σ0)e−bz when

σ0 > σ1, σ1 = 0.5 and σ0 = 1.5 (S/m), the numerical solutions of the magneticfield at each node is calculated by using Matlab program as shown in Figure.

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Graphs of the relationship between magnetic field and distance of receiver fromsource at various depths.

(a) (b) (c)

(d) (e)

Figure 3.4: The relationship between magnetic field and distance of receiverfrom source at various depths when (a) b = 0.01 m−1 (b) b = 0.05 m−1 (c) b= 0.1 m−1 (d) b = 0.2 m−1 (e) b = 0.3 m−1

From Figure 3.4(a) to (e), when b = 0.01, 0.05, 0.1, 0.2 and 0.3 m−1, respec-tively, we can see that the value of magnetic field decreases exponentially as rincreases and it decreases as z increases as well. The value of magnetic fieldincreases when b increases and it decreases slowly from b = 0.2 m−1 to b = 0.3m−1. This results is caused by the conductive of the ground and the verticallylocation of its which is as same as the case of decreasing conductivity profile.

(a) (b) (c) (d)

Figure 3.5: The relationship between magnetic field and distance of receiverfrom source when b varies from 0.01, 0.05, 0.1, 0.2 and 0.3 m−1 and z is fixed.(a) z=20 m (b) z=60 m (c) z=100 m (d) z=140 m

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Figure 3.5(a) to (d)represents the values of magnetic field which are plottedagainst r when b varies and z is fixed at 20, 60, 100 and 140 m, respectively.We can see that the value of magnetic field where b = 0.01, 0.05, 0.1, 0.2 and0.3 m−1 decreases exponentially as r increases and it has similar values whenz increases because the value of magnetic field decreases to zero and has valuenear zero when z increases as b varies. From Figure 3.5(b), (c) and (d), thethree curves representing the values of magnetic field which are plotted againstr when b = 0.1, 0.2 and b = 0.3 m−1 have a different behavior from the others.The value of magnetic field is smaller than that when b = 0.2 m−1, as we cansee the curve cross the line for b = 0.2 m−1. These results occur according tothe conductive of ground and the spacing distance between source and receiver.

Contour graphs of the relationship between magnetic field and distance ofreceiver from source at various depths.

(a) (b) (c)

(d) (e)

Figure 3.6: Contour graphs of magnetic field at different distances of receiverfrom source and different depths when (a) b = 0.01 m−1 (b) b = 0.05 m−1

(c) b = 0.1 m−1 (d) b = 0.2 m−1 (e) b = 0.3 m−1

From Figure 3.6(a) to (e), when b = 0.01, 0.05, 0.1, 0.2 and 0.3 m−1, thered color shows the area when the value of magnetic field is high and the bluecolor shows the area when the value of magnetic field is low. The value ofmagnetic field increases when b increases and it decreases slowly from b = 0.2m−1 to b = 0.3 m−1, as we can see in Figure 3.6(a) to (e).

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4 Conclusions

A mathematical model of the magnetic fields is conducted by using partialdifferential equations. Numerical solutions of partial differential equations arecomputed by using finite element method (FEM) to find the value of the mag-netic field at various locations of the ground. In our models, the ground have anexponentially decreasing and increasing conductivity profiles. The Galerkin’sMethod of Weighted Residuals is applied in finite element method. There isa source providing a DC voltage and a receiver on the ground surface whichpicks up the signal from r = 10 m to r = 190 m. We begin our study withthe relationship between magnetic and electric fields. The vector equationsfor computing magnetic and electric field is very difficult. Thus, in most re-search, we change the vector equations into scalar equations by considering thecomponents of the vector. We finally arrive with partial differential equationdescribing the system and it is surprisingly that our equation is independentfrom the electric current. The numerical results is calculated and plotted byusing MATLAB R2008b program to show the behavior of magnetic field atdifferent depths and distances from source point. In our research, the behav-ior of magnetic field decreases to zero when the depth of soil increases. As wellas the case of increasing the space between source-receiver, the magnetic fielddecreases to zero too. The value of b is an important role for the conductionof ground and effect to the magnetic field quantities as well. The comparisionof the quantities of magnetic field for the case of σ as an increasing function ishigher than the case of σ as a decreasing function according to the advantage ofthe DC source that better reflex at very high depth than on the ground surface.

Acknowledgements. The authors would like to thank the Departmentof Mathematics, Faculty of Science, Silpakorn University and Centre of Excel-lence in Mathematics for continuous financial and equipments support.

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Received: February 23, 2015; Published: March 27, 2015

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VVol. 9, no. 49-52, 2015 doi:10.12988/ams ISSN 1312-885X (Print) ISSN 1314-7552 (Online)

APPLIED MATHEMATICAL SCIENCES

Journal for Theory and Applications

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Applied Mathematical Sciences, Vol. 9, 2015, no. 49 – 52

Contents

U. U. Abylkayrov, S. E. Aitzhanov, L. K. Zhapsarbayeva, Solvability of the inverse problem for a heat convection system with integral condition of overdetermination 2403

Jaegwi Go, Je-Hyun Lee, Temperature distribution profiles during vertical continuous casting process 2423

Asmala Ahmad, Shaun Quegan, The effects of haze on the accuracy of satellite land cover classification 2433

Bashar Zogheib, Ali Elsaheli, Approximations to the t distribution 2445

R. Harikumar, P. Sunil Kumar, Frequency behaviours of electro- encephalography signals in epileptic patients from a wavelet thresholding perspective 2451

Sergei Soldatenko, Rafael Yusupov, Shadowing property of coupled nonlinear dynamical system 2459

Andrea Colantoni, Lavinia M. P. Delfanti, Filippo Cossio, Benedetto Baciotti, Luca Salvati, Luigi Perini, Richard Lord, Soil aridity under climate change and implications for agriculture in Italy 2467

Elmanani Simamora, Subanar, Sri Haryatmi Kartiko, Asymptotic property of semiparametric bootstrapping kriging variance in deterministic simulation 2477

M. Ahsanullah, M. Shakil, B. M. Golam Kibria, F. George, Distribution of the product of Bessel distribution of first kind and gamma distribution- properties and characterization 2493

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PPonco Siwindarto, ING. Wardana, M. Rasjad Indra, M. Aris Widodo, Sudden cardiac death prediction using Poincare plot of RR interval differences (PORRID) 2515

K. Ananthi, J. Ravi Sankar, N. Selvi, Bipartite universal domination of zero divisor graph 2525

Daya K. Nagar, Danilo Bedoya-Valencia, Arjun K. Gupta, Bivariate generalization of the Gauss hypergeometric distribution 2531

C. S. Ryoo, Numerical investigation of the zeros of q-extension of tangent polynomials 2553

Embay Rohaeti, Sri Wardatun, Ani Andriyati, Stability analysis model of spreading and controlling of tuberculosis (case study: tuberculosis in Bogor region, West Java, Indonesia) 2559

Pablo Soto-Quiros, Application of block matrix theory to obtain the inverse transform of the vector-valued DFT 2567

Priyanuch Tunnurak, Nairat Kanyamee, Suabsagun Yooyuanyong, Finite element magnetic field response of an exponential conductivity ground profile 2579

Jerico B. Bacani, Julius Fergy T. Rabago, On linear recursive sequences with coefficients in arithmetic-geometric progressions 2595

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182

Biography Name Miss Priyanuch Tunnurak Address 496 Moo 10 Road Nongkhaem, Khet Nongkhaem, Bangkok, 10160 Date of Birth 16 May 1989 Education 2010 Bachelor of Science in Applied Mathematics, Silpakorn University 2014 Master of Science in Mathematics, Silpakorn University

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