Eindhoven University of Technology MASTER The impressed

Eindhoven University of Technology

MASTER

The impressed current cathodic protection system

Kakuba, G.

Award date:2005

Link to publication

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https://research.tue.nl/en/studentTheses/274bc367-4285-40af-9263-ba842f0700e1

TECHNISCHE UNIVERSITEIT EINDHOVEN

Department of Mathematics and Computer Science

MASTER’S THESIS

The Impressed Current Cathodic

Protection System

by

Godwin Kakuba

Supervisors:

Prof. Dr. R. M. M. Mattheij (TUE)Dr. Ir. M. J. H. Anthonissen (TUE)

Dr. E. S. A. M. Lepelaars (TNO)

Eindhoven, August 2005

Acknowledgements

During the evolution of this thesis, I enjoyed not only the immense expertisebut also the enormous experience of my supervisors Prof. Dr. R. M. M. Mattheij,Dr. Ir. M. J. H. Anthonissen and Dr. E. S. A. M. Lepelaars. I would like tothank them for always having time to answer my questions, big or small, evenwithout making appointments in advance. Through their own research, com-ments, recommendations and questions, they have encouraged, supported andenlightened me. I benefited loads from Prof Mattheij’s and Dr. Anthonissen’smastery of Numerical Analysis, they always steered me on track whenever myefforts tended to flare out of course. Prof Mattheij’s humorous jokes during ourmeetings offered great lubrication between life and academics. Dr. Lepelaars’authority in electromagnetics and Fortran was vital in the production of thisthesis. I always learned a lot from our meetings at the company in The Haguewhere I also enjoyed those coin free soup machines. I am also indebted to Dr.Lepelaars for allowing me to directly use part of his research work in this thesis.

I owe a great deal of gratitude to colleagues, friends and members of my familyfor their profound moral support. Thanks to all my coursemates and friendsat TU/e for the jovial time we spent together. I am also thankful to membersof Makerere University staff studying abroad who, through our kimeeza mailinglist, always posted fun and messages of encouragement among other things. Iliked this one: ‘‘Research is like a pregnancy, it gets bitter when you are about todeliver.”

This project necessitated a good knowledge of integral equation methods andthe BEM in particular, a subject that was virgin to me. Building on the confidenceand knowledge I obtained from the first-year lectures, I have acquired goodtraining in the area of the BEM as a result of this project. I also had a closer lookinto the area of potential theory. Save for the basic electrolysis knowledge that Ihad, this was my introduction into the field of cathodic protection systems. Thiswas also my first project ever in Fortran. What a learning experience!

Contents

Acknowledgements iii

Contents vii

1 Introduction 1

1.1 Background . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.2 Outline of this thesis . . . . . . . . . . . . . . . . . . . . . . . . . . 5

2 The Impressed Current Cathodic Protection System 7

2.1 Basic electrochemistry . . . . . . . . . . . . . . . . . . . . . . . . . 10

2.2 Electrode kinetics . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

2.3 Passive and active cathodic protection . . . . . . . . . . . . . . . . 13

2.4 Boundary conditions . . . . . . . . . . . . . . . . . . . . . . . . . . 15

2.5 The model problem . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

3 Potential theory and the Boundary Element Method 21

3.1 Mathematical background . . . . . . . . . . . . . . . . . . . . . . . 21

vi

3.1.1 Green’s Theorems . . . . . . . . . . . . . . . . . . . . . . . 23

3.1.2 Fundamental solution of the Laplace equation . . . . . . . 24

3.1.3 Integral properties of Green’s functions . . . . . . . . . . . 27

3.2 Integral formulation of the Laplace equation . . . . . . . . . . . . 28

3.2.1 Basic integral equation: Internal points . . . . . . . . . . . 28

3.2.2 Basic integral equation: Boundary points . . . . . . . . . . 31

3.2.3 Infinite regions . . . . . . . . . . . . . . . . . . . . . . . . . 35

3.2.4 Physical interpretation of single and double layer potentials 42

3.2.5 The electric field intensity . . . . . . . . . . . . . . . . . . . 44

3.3 The BEM for the Laplace equation . . . . . . . . . . . . . . . . . . 47

3.3.1 Constant elements . . . . . . . . . . . . . . . . . . . . . . . 47

3.3.2 Calculation of the matrix elements . . . . . . . . . . . . . . 51

4 Numerical results and discussion 57

4.1 An example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

4.2 Solution to the model problem . . . . . . . . . . . . . . . . . . . . 61

4.3 Current flow out of the box . . . . . . . . . . . . . . . . . . . . . . 68

4.4 Examples for other electrode positions . . . . . . . . . . . . . . . . 72

4.5 Richardson extrapolation . . . . . . . . . . . . . . . . . . . . . . . . 73

5 Conclusions and Recommendations 77

5.1 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77

vii

5.2 Recommendations . . . . . . . . . . . . . . . . . . . . . . . . . . . 78

Appendix 81

A Gauss integration for triangular domains 81

Bibliography 87

Chapter 1

Introduction

1.1 Background

Corrosion is a process in which metals have a natural tendency to return to theiroxidised forms. It is an electrochemical oxidation reaction that requires an an-ode, a cathode, an electron pathway and an electrolyte for it to take place. Thecomplete system of these components is called a corrosion cell. For marine sub-merged objects, seawater is the electrolyte. The process is greatly acceleratedby increase in temperature, oxygen, water speed and the presence of chloridesand sulfides. This explains the severe corrosion in the splash zone of water frontstructures. The process of corrosion is mostly a nuisance as it leads to the dam-aging of metals and must be prevented from taking place. This is often thecase in buried metallic structures such as pipelines, oil and gas wells, offshorestructures, seagoing ship hulls, marine pilings, water tanks and some chemicalequipments. The most common methods to prevent corrosion are by;

1. preventing access of the electrolyte,

2. reversing the flow of electrons,

3. corrosion resistant alloys like in stainless steel where chromium and nickelprotect the steel from corrosion allowing it to be used in most kitchen uten-sils and by

4. corrosion allowance.

2 Chapter 1. Introduction

The first method involves applying a coating on the surface to be protected thatprevents the electrolyte from reaching it. This may be organic coating such as inpainting or metallic coating such as in galvanizing. The second method is com-monly known as cathodic protection and is widely used in underground andseawater structures. The third method is the use of alloys that do not corrodein these environments. The fourth method allows the corrosion to proceed andincorporates enough structural material in the design to last for the intendedservice life. The last two are more expensive methods to opt for.

The focus of this thesis is cathodic protection. Cathodic protection is a tech-nique used to prevent the corrosion of a metal by making the metal a cathode.It is commonly used in the protection of the exterior surfaces of pipelines, shipshulls, storage tank bases, jetties and harbour structures, floating and sub seastructures, etc. Cathodic protection is also used to protect the internal surfacesof large diameter pipelines, ships’ tanks, storage tanks for oil and water andwater-circulating systems.

The first reported practical use of cathodic protection is generally credited to SirHumphrey Davy in the 1820s. Davy’s advice was sought by the British RoyalNavy in investigating the corrosion of copper sheeting used for cladding thehulls of naval vessels. Davy found that he could preserve copper in seawaterby the attachment of small quantities of iron, zinc or tin. The copper became,as Davy put it, cathodically protected. It was quickly abandoned because byprotecting the copper its antifouling properties became retarded, hence reducingthe streamline of the ships, as they began to collect marine growths [Kean andDavies, 2003].

The most rapid development of cathodic protection was made in the UnitedStates of America and by 1945, the method was well established to meet the re-quirements of the rapidly expanding oil and natural gas industry, which wantedto benefit from the advantages of using thin-walled steel pipes for undergroundtransmission [Kean and Davies, 2003]. In the United Kingdom, where low-pressure, thicker-walled castiron pipes were used extensively, very little cathodicprotection was applied until the early 1950s [Kean and Davies, 2003]. Since the1950s the method has become popular and is now well established and widelyused.

The processes involved in modern day cathodic protection systems are quitecomplex. To effectively build a cathodic protection system, the chemical pro-cesses that involve polarisation of the electrodes must be taken into account.The rate of corrosion will be affected by resistance to (ionic) current flow of theelectrolyte, resistance to current flow (by electron conduction) in the conducting

1.1 Background 3

materials and in the connection between them. Furthermore, the resistance as-sociated with polarisation behaviour of the anode and the resistance associatedwith polarisation behaviour of the cathode (cathode efficiency) are important.These factors are usually given by polarisation curves and they take into accountthe chemical properties of the metals concerned. In a mathematical languagethese factors define our boundary conditions. The geometry of the structuresto be protected are usually highly irregular. Yet a proper distribution of anodesmust be established to ensure optimum protection but still minimize costs. InFigure 1.1, we show an example of a typical ship and an anode surface. To op-timise the cathodic protection system, one will also need to know the potentialand current profiles due to a cathodic protection system. A lot of research isgoing on to calculate the potential and electric field distribution while takinginto account all the above factors as in [Sun and Liu, 2000, Rikte, 2003, McCaf-ferty, 1977]. Knowledge of the electric field and potential distributions due to

(a) (b)

Figure 1.1: A photo of a typical ship (under construction) in which ca-thodic protection is necessary, (a), the isolated eye in (b) is a typicalanode on that ship.

cathodic protection can be used to establish an optimal potential profile for theship, whereby critical areas are protected under all conditions. The electric fielddistribution can also be used in assessing the quality of the coating in ships; theelectric field will, for example, be strong around highly conducting parts of theship.

The goal of this project is to determine the electric field distribution and poten-tial distribution around a cathodically protected surface of a ship. This surfaceconsists of four parts; the anode with an insulating shield around it, the coatedhull and the propeller surfaces, see Figure 2.1. The coated hull and the propeller


surfaces have different polarisaton curves and are the surfaces to be protectedfrom corrosion. The potential outside the surface satisfies the Laplace equation.

In this thesis we solve a model problem where by the ship geometry is sim-plified to a simpler shape, a cuboid. We consider sea water as a continuumwith uniform body properties. Most of the ship’s surface is assumed to be per-fectly insulating apart from the cathode and anode surfaces where we assumelinear Dirichlet boundary conditions. The resulting mathematical problem is aso-called external potential problem. Together with the boundary conditions, wehave a boundary value problem which we will formulate as an integral equationand solve numerically. One efficient method for solving such a problem is theBoundary Element Method. Since the domain of our problem is the bulky infi-nite region outside the ship, the main advantage of using the Boundary ElementMethod is that only the boundary of the domain requires discretisation. Thusthe dimension of the problem is effectively reduced by one. This is in contrastwith other common numerical methods such as the finite element method andthe finite difference method where the whole domain requires discretisation. Weshall therefore replace the Laplace equation defined in the domain by an integralequation valid on the boundary surface. Discretisation of this equation and ap-plication of boundary conditions will result in an algebraic system of equationsthat needs to be solved. For linear boundary conditions the resulting systemis linear and can be solved using an LU decomposition. In case of nonlinearboundary conditions the resulting system is nonlinear and an iterative methodhas to be employed.

Until the beginning of the eighties, the BEM was known as Boundary IntegralEquation Method (BIEM). It has its origin in the work of George Green (1793 -1841). In 1828 he formulated the integral representation of the solution for theDirichlet and Neumann problems of the Laplace equation by introducing theso-called Green’s function for these problems [Katsikadelis, 2002]. In 1872 Betti(1823 - 1892) presented a general method for integrating the equations of elastic-ity and deriving their solution in integral form. In 1885, Somigliana used Betti’sreciprocal theorem to derive the integral representation of the solution for theelasticity problem. However Fredholm was the first one to use singular bound-ary integral equations to find the unknown boundary quantities for problems ofpotential theory.

Boundary integral equation methods (BIEM) are classified into the direct andthe indirect integral equations. The direct boundary integral equation method isderived from the fact that once the fundamental solution of a given equation to-gether with the prescribed boundary conditions on a geometrically well-definedboundary is known, then the solution of such a boundary value problem is also

1.2 Outline of this thesis 5

known in the form of an integral equation and can be computed numerically.The integral equation is derived through the application of Green’s second the-orem. The indirect method is based on the assumption that the solution can beexpressed in terms of a source density function defined on the boundary. Ineither case, the use of the fundamental solution reduces the dimension of theproblem by unity whereby the volume integrals reduce to surface integrals andthe surface integrals reduce to line integrals and the governing differential equa-tions to integral equations.

The BIEM became popular in the period from 1960 through 1975. M. A. Jawson(1963), G. T. Symm (1963) and T. A. Cruise and F. J. Rizzo (1968) are the fore-runners of what was to be called the BEM in 1978 at the second conference atSouthampton, UK, and on the publication of the first monograph-style text onthe BEM by C. A. Brebbia (1978) [Kythe, 1995]. Since then the research groupsheaded by C. A. Brebbia, P. K. Banerjee, R. P. Shaw, J. A. Liggett, H. Antes andothers, have advanced the subject to a high degree of adaptability in numericalsolution of boundary value problems.

1.2 Outline of this thesis

In Chapter 2 we present a deeper discussion of cathodic protection. As alreadymentioned in the previous section, corrosion is an electrochemical reaction. InChapter 2, basic electrochemistry is discussed to properly understand the cor-rosion reaction. A clear distinction between passive cathodic protection whichinvolves the use of sacrificial anodes and active cathodic protection which in-volves the use of an impressed current is made. The chemical processes whichdetermine the rate of corrosion are discussed and polarisation curves are pre-sented. These curves describe different types of boundary conditions which arealso discussed. In the last section of this chapter the model problem is presented.The model comes from simplification of the geometrical shape of the ship andmaking a choice of the boundary conditions to be used. For the model, we shallmake assumptions on the chemical properties of sea water. We finally end upwith an exterior potential problem with linear boundary conditions.

In Chapter 3 the methodology is developed. The boundary element method(BEM) will be introduced and used to solve our model problem. In the BEM,once we know the boundary distribution of the unknown function and its deriva-tives then the solution at any point in the domain can be calculated. It is this re-duction in dimension that makes the BEM a proper choice for problems defined


over large domains. For linear and elliptic problems, the BEM is vastly superiorin efficiency and accuracy to FEM and FDM [Pozrikidis, 2002]. Problems thatcan be solved on a laptop computer using the BEM may require the use of a su-percomputer by finite difference and finite element methods for the same levelof accuracy.

Since our problem is a potential problem, basic knowledge of the fundamen-tal principles underlying potential theory and the integral equation representa-tions is imperative. In Chapter 3 we discuss elements of potential theory thatare crucial for our solution method. The derivation of the integral equation foran exterior problem follows from that of an interior problem. For this reason,we discuss the derivation of integral equations for an interior potential problemand then show how we get the integral equation for the exterior problem. Thenthe discretisation of the integral equation is discussed. Formation of the linearsystem will involve integration of the fundamental solution or its derivatives.Gaussian quadrature rules are used on elements where the integrand is nonsin-gular and analytical solutions are used for the elements where the integrandsare singular.

In Chapter 4, our simulation results are presented and discussed. To verify ournumerical algorithm as well as its implementation, an example is discussed. Forthis example, we know the exact solution and we show the accuracy of the BEMby comparing exact and numerical values. Then the results of our model prob-lem are presented. The results show that the potential goes to zero monotoni-cally away from the ship. We also experiment on other positions for the anodesand our results show similar trends. As a representation of the current we cal-culate the electric field intensity in various planes in space. We also presentprojections of the field in different xy-planes.

For the interests of the reader an appendix is attached to explain the triangularcoordinates used in the Gaussian quadrature rules for a triangle.

Chapter 2

The Impressed Current CathodicProtection System

The cathodic protection system, or ICCP (Impressed Current Cathodic Protec-tion) System, is a system to prevent bare steel surfaces from corroding.

Figure 2.1: Part of a ship’s surface showing atypical anode and a propeller shaft, one of themost vulnerable parts of the ship.

Corrosion is an electrochemical process that involves passage of electrons from

8 Chapter 2. The Impressed Current Cathodic Protection System

one substance, usually metallic, called the anode to another substance called thecathode. It is at the anode that the oxidation reaction which is responsible forthe corrosion of the metal involved takes place according to the following ionicequation, in the case of steel:

Fe(s) → Fe2+(aq)

+ 2e−, (2.1)

where Fe represents a ferrous atom, Fe2+ a ferrous II ion, e an electron and thesubscripts s and aq imply the solid state and the aqueous state respectively.The electrons so produced pass on to the cathode where they are used up in areduction reaction:

2H+(aq)

+ 2e− → H2(g) (2.2)

in acidic solutions, or:

O2(g) + 2H2O(l) + 4e− → 4OH−(aq)

(2.3)

in neutral solutions, see Figure 2.2, metal II is a metal that is more electropositivethan steel.

Anode Cathode(Steel) (Metal II)

(corrosion)

electron flow

current flow

electrolyte

2Fe → 2Fe2+ + 4e− O2 + 2H2O + 4e− → 4OH−

Figure 2.2: Bimetallic corrosion cell.

In equations (2.2) and (2.3), H+ represents a hydrogen ion, O2 oxygen gas, H2O a

9

water molecule and OH− a hydroxyl ion. The subscripts l and g imply the liquidstate and the gaseous state respectively. Therefore corrosion basically occurs atthe anode. Cathodic protection involves attaching an anode to the surface to beprotected and supplying a direct current through it such that all the other partsof the surface become cathodic and therefore do not corrode, see Figure 2.3.

Figure 2.3: The ICCP system.

Cathodic protection can be achieved in two ways, viz:

(i) By the use of sacrificial anodes, or

(ii) By an impressed current.

In (i), a reactive auxiliary metal is directly attached to the metal to be protectedas an anode. The difference in the natural potentials between the anode and thesteel as indicated by their relative positions in the electrochemical series causes acurrent to flow through the electrolyte from the anode to the steel. In (ii) an inertanode is attached to the steel surface and an external source of direct currentis used to impress a current from the anode onto the cathode surface hence thename impressed current cathodic protection system. This is the system depicted inFigure 2.3 and is the one with which we are concerned in this thesis.


In Sections 2.1 to 2.4 we give details of the corrosion processes for steel surfacesin seawater and cathodic protection. In these sections, the material has beenmainly drawn from [Lepelaars, 2005] with a few additions or subtractions.

2.1 Basic electrochemistry

To better understand how the ICCP works, we study the corrosion process inmore detail. Consider an electrolytic cell which consists of two glass jars con-

ZnCl2

Zn

CuSO4

Cu

Salt bridge

V0

Anode Cathode

Figure 2.4: Galvanic or electrolytic cell to illustrate the corrosion process.

taining a solution of zinc chloride (ZnCl2) and copper sulphate (CuSO4) respec-tively. A zinc rod is immersed in the zinc chloride solution and a copper rodin the copper sulphate solution and the two are connected to a battery with im-pressed voltage V0 = 0 via an external wire. A salt bridge is used to complete theelectric circuit from one glass to the other without disturbing the ion concentra-tions of both glasses, see Figure 2.4. Further we assume standard conditions, i.e.,both the CuSO4 and the ZnCl2 solutions have a concentration of 1 mole metalions per liter (1.0 M). At the anode, an oxidation reaction occurs:

Zn(s) → Zn2+(aq)

+ 2e−, (2.4)

with a standard potential of −0.76 V. Here Zn represents a zinc atom and Zn2+

a zinc ion. At the cathode, a reduction reaction takes place:

Cu2+(aq)

+ 2e− → Cu(s), (2.5)

2.1 Basic electrochemistry 11

with a standard potential of +0.34 V. The standard potential difference understandard circumstances is defined as

E0 = Ecathode − Eanode = 0.34 + 0.76 = 1.10 V. (2.6)

In the case of a voltage V0 = 0 we see that the potential of the cathode is higherthan the potential of the anode. Hence, through the wire, an electric currentflows from the cathode to the anode. Because of conservation of electric charge,the same electric current must flow through the salt bridge, flowing from theZnCl2 solution to the CuSO4 solution.

What happens if we do not consider standard conditions, i.e., if the concen-trations of both solutions are different? Then the potential difference betweencathode and anode is given by

Ecathode − Eanode = E0 −RT

nFln Q, (2.7)

which is called the Nernst law. In this formula, E0 is the standard potentialdifference, R is the gas constant, T is the absolute temperature in Kelvin, n is thenumber of electrons involved in the reaction, F is Faraday’s constant (9.65 × 104

C/mole) and Q is the equilibrium constant. The latter is in this case defined as

Q =

[

Zn2+]

anode[

Cu2+]

cathode

, (2.8)

here the square brackets denote the ion concentration of the enclosed quantity.Under standard conditions,

[

Zn2+]

anode=[

Fe2+]

cathode= 1 mole/l and thus

ln Q = 0. In the case that Ecathode − Eanode > 0, a spontaneous reaction takesplace and we call the cell of Figure 2.4 a Galvanic cell. The zinc anode gradu-ally dissolves while copper precipitates on the cathode. The equilibrium can beinfluenced by introducing an external voltage source as indicated in Figure 2.4.The Nernst equation then gets an extra term:

Ecathode − Eanode = E0 − V0 −RT

nFln Q. (2.9)

By choosing V0 large enough, we can achieve that Ecathode − Eanode < 0. The cellis then called an electrolytic cell. The external voltage then enforces the electriccurrent to flow opposite of the ‘natural’ direction. The copper dissolves whilezinc precipitates on the anode. Equilibrium is obtained when Ecathode − Eanode =

0. In this situation no electric current flows; it occurs when V0 is chosen as

V0 = E0 −RT

nFln Q, (2.10)


which is called the zero-current cell potential. As we have seen, whether or nota reaction will occur follows from the basics of electrochemistry. It does not tellat what speed the reaction takes place. We will come to that in the followingsection.

2.2 Electrode kinetics

In this section we discuss the reaction speed, or better, the rate of electron trans-fer. Let Ox represent an oxidising agent and Red a reducing agent. We considerthe half reaction

Ox + ne− Red, (2.11)

expressing that n electrons are transferred during a reduction/oxidation step.The total current that flows in the half reaction depends on both partial anodicand cathodic current densities Ja and Jc, i.e.,

J = Ja + Jc = nFAkred [Ox]0 − nFAkox [Red]0 . (2.12)

In this equation, A represents the surface area, F is Faraday’s constant, kred andkox are the rate constants of the reduction and oxidation and [Ox]0 and [Red]0are the surface concentrations. It can be shown that kred and kox take the form

kred = Ared exp

[

−αnF(Φ − Φe)

RT

]

, kox = Aox exp

[

(1 − α)nF(Φ − Φe)

RT

]

,

(2.13)

where Ared, Aox and α are some constants and where Φe is the voltage at equi-librium. Typically α = 0.5 which indicates that the transition state responds toa potential difference in a manner half way between the reactants and the prod-ucts response. These relations are used to arrive at the so-called Butler-Volmeror Tafel equation:

J = J0

[Ox]0

[Ox]bulk

exp

[

−αnF(Φ − Φe)

RT

]

−[Red]0

[Red]bulk

exp

[

(1 − α)nF(Φ − Φe)

RT

], (2.14)

where J0 is the exchange current density. If we assume that the solution is wellmixed, the bulk and surface concentrations may be assumed to be identical, sim-plifying the Butler-Volmer equation to

J = J0

exp

[

−αnF(Φ − Φe)

RT

]

− exp

[

(1 − α)nF(Φ − Φe)

RT

]. (2.15)

2.3 Passive and active cathodic protection 13

This is in fact a simple polarisation function. It is clear that if Φ = Φe, thenno current flows. In the particular case of steel in seawater, the general form ofthe cathodic polarisation function requires modification, see [Sun and Liu, 2000]and [Bucaille and Warner, 1997]:

Jc = J0(i1 − i2 − i3), (2.16a)

i1 = exp[(Φ − Φ1)/b1], (2.16b)

i2 =

(

1

iL+

1

exp[(Φ − Φ2)/b2]

)−1

, (2.16c)

i3 = exp[−(Φ − Φ3)/b3], (2.16d)

where Φi and bi, i = 1, 2, 3 and iL are constants and where i1, i2 and i3 representthe current density generated by iron oxidation, oxygen reduction and hydro-gen evolution, respectively. The six unknown parameters may be fitted fromexperimental data. An interesting collection of literature on experimentally de-termined polarisation curves can be found in [Stefec, 1990], which is based onthe PolarBank. The PolarBank is an initiative in the early 90’s to gather polarisa-tion data in a database so that it is available to everyone.

To conclude this section, electrode kinetics gives a relation between the currentdensity at the electrode surface and the potential difference, i.e., it deals aboutreaction velocity.

2.3 Passive and active cathodic protection

Suppose that a lump of iron is immerged into seawater. The following two reac-tions take place at the surface:

Fe → Fe2+ + 2e−, O2 + 2H2O + 4e− → 4OH−. (2.17)

At the iron surface an electrochemical double layer develops, where the actualdissolution takes place. This layer has a thickness of only several atoms, wherea very strong potential difference is created. The potential of seawater is definedas 0 V. The potential of the steel is about −600 mV. This value is the result of anequilibrium between both half reactions, as follows from the schematic diagramin Figure 2.5. For a potential difference of −800 mV, the iron dissolution is inhib-ited, but the half reaction of O2 is promoted. Both half reactions are, however,connected by the law of conservation of charge. This situation is also shown bythe Butler-Volmer equation. Now if we would trade the lump of iron for a lump


−600

−1000

Fe → Fe2+ + 2e−

Zn → Zn2+ + 2e−

O2 + 2H2O + 4e− → 4OH−

∆Φ

(mV

)

log(

|i|

µA/cm2

)

Figure 2.5: Tafel plot representing the relation between potential difference andelectric current for some half reactions.

of zinc, the situation would in essence still be the same, except for a lower equi-librium voltage Φe of about −1000 mV. Next we consider the situation of a lumpof zinc which is galvanicly connected to a lump of iron in seawater. Both met-als are good conductors, hence they will have the same intermediate potentialof, say, −950 mV. Because of this new equilibrium value, the dissolution of ironis inhibited but the dissolution of zinc is promoted. This principle is used forpassive cathodic protection. The zinc anode is sacrificed to save the iron lump.As an alternative to keep the potential of the iron lump substantially low, thezinc could be replaced by an active anode, i.e., a platinum anode with a voltagesource, as indicated in Figure 2.6. This process is called active cathodic protec-tion. At the platinum anode, the following reaction takes place:

2Cl− → Cl2 + 2e−. (2.18)

In practice, the steel hull is maintained at a potential between −800 and −1000

mV. Exceeding the latter limit results in a damage of the coating. In this thesis weshall consider active cathodic protection where we have an anode and a cathodemaintained at certain potential values.

2.4 Boundary conditions 15

−950 mV FeZn

O2 + 2H2O + 4e− → 4OH−Zn → Zn2+ + 2e−

−950 mV

FePt

O2 + 2H2O + 4e− → 4OH−2Cl− → Cl2 + 2e−

Figure 2.6: Illustration of the principle of passive (top) and active (bottom) ca-thodic protection.

2.4 Boundary conditions

If we want to know the electric currents that are related to the chemical processesof active cathodic protection, it is necessary to describe the processes in termsof electromagnetic quantities. If we restrict to static currents, the problem canbe described by an electric potential. For the schematic situation as depictedin Figure 2.7, the potential u satisfies Laplace’s equation in the seawater layer,∇2u = 0. Further, this potential satisfies boundary conditions at the

• anode surface, Γ1

• cathode surface, Γ2

• insulating hull surface, Γ3,


• the seawater-seabed interface, Γ4,

• the seawater-air interface, Γ5.

air

Γ3

Γ5

seawater, σ ∇2u = 0

J = σE

Γ4

seabed

Anode

Γ1

Cathode

Γ2

Insulating

surface

Figure 2.7: Schematic view of the potential problem related to activecathodic protection.

These boundary conditions can be derived if we take into account the electricfield E and the electric current density J inside the seawater, which are relatedby

E = −∇u, J = σE, (2.19)

where σ represents the seawater conductivity. In this thesis we have considereda case where a ship model is completely immersed in seawater which is consid-ered to be a homogeneous medium. Thus we shall only focus on the other threesurfaces and leave out the last two. For the insulating hull surface,

n · J = σn · E = −σn · ∇u.

Since no current flows through the insulating hull, this results in the condition:

∂u

∂n= n · ∇u = 0 at the insulating hull surface, (2.20)

where n is the outward pointing normal vector. The left-hand side is referredto as the normal derivative of the potential. At the cathode surface, the rela-tion between current density and potential difference is given by a polarisationcurve:

J · n = −σ∂u

∂n= fc(u), (2.21)

2.5 The model problem 17

with the general shape of fc(u) given by equations (2.16). At the anode surface,the contact with the seawater is extremely well. Therefore either of the followingrelations may be assumed at the anode surface:

u = ua, (2.22a)

σn · ∇u = −I

A, (2.22b)

J · n = −σ∂u

∂n= fa(u). (2.22c)

The first boundary condition expresses that each point of the anode surface hasthe same potential. The second boundary condition expresses a homogeneouslydistributed current density flowing out of the anode surface. Here I is the totalcurrent flowing through the anode surface and A is the total surface area. Thethird condition, is a more general polarisation curve for anodic surfaces, usedby [Bucaille and Warner, 1997, McCafferty, 1977, Morris and Smyrl, 1988, Doigand Flewitt, 1979]. Boundary conditions involving polarisation curves are non-linear, i.e., the relation between the potential and the current density is non-linear. For example, Figure 2.8 shows polarisation curves for copper, steel andzinc. In some references, the polarisation curve is linearised [Morris and Smyrl,1989, McCafferty, 1977]. For our model, we shall use (2.22a).

2.5 The model problem

In this section we present a simplified model of the ICCP on which we havebased our solution in this thesis. The following simplifications have been madeto obtain the model:

• For the shape of the ship we shall consider a cuboid. The shapes of theelectrodes are also assumed rectangular.

• For the domain, instead of considering the three layers of air, seawaterand the seabed as depicted in Figure 2.7, we shall have our box totallyimmersed in seawater as depicted in Figure 2.9.

• For the boundary conditions we shall consider linear boundary conditionsat the electrodes, that is, a potential is prescribed at the cathode and theanode as in equation (2.22a). In addition, the rest of the surface is assumedto be a perfect insulator.


Figure 2.8: Polarisation curves for copper steel and zinc.

2.5 The model problem 19

Cat

hode

Anode

Seawater

Figure 2.9: The simplified model.

With these simplifications, we seek the distribution of the potential and the cur-rent everywhere outside the box but in seawater. Thus: Let Ω be the regionoutside the box, Γ1 the anode surface, Γ2 the cathode surface and Γ3 the insulat-ing hull surface. Let u be the potential outside the box and r the position vectorof a point in space, then we have the following exterior potential problem:

∇2u(r) = 0 r ∈ Ω,

u(r) = ua r ∈ Γ1,

u(r) = uc r ∈ Γ2,

∂u

∂n(r) = 0 r ∈ Γ3.

(2.23)

where n(r) is the unit normal vector at r and∂u

∂n(r) = n(r) · ∇u(r). For the

solution of this problem, we shall, in Chapter 3, devote our attention to potentialtheory and explain the solution method used to solve the potential problem. Theresults are discussed in Chapter 4.

Chapter 3

Potential theory and theBoundary Element Method

In this chapter, we give an insight into potential theory and derive some funda-mental identities used to solve the potential problem presented in Chapter 2.

3.1 Mathematical background

In this section we discuss the mathematical concepts that are vital for the de-velopment of the Boundary Element Method which is used to solve the poten-tial problem discussed in Chapter 2. The Boundary Element Method (BEM) isdeveloped for the solution of engineering problems described by the potentialequation

∇2u(r) = f(r), r ∈ Ω. (3.1)

This is the governing equation for potential theory. When f(r) = 0 it is knownas the Laplace equation whereas when f(r) 6= 0 it is known as the Poisson equa-tion. Its solution u = u(r) represents the potential produced at a point r in thedomain Ω due to a source f(r) distributed over Ω. The solution of (3.1) is soughtin a closed domain Ω having a boundary ∂Ω on which either the function u

or its derivative ∂u/∂n, where n is the unit outward normal vector to ∂Ω, isprescribed as shown in Figure 3.1.

22 Chapter 3. Potential theory and the Boundary Element Method

Ω

n

∂Ω

Ωc

Figure 3.1: Definition of terms for the potentialproblems given in (3.2) to (3.7).

The boundary value problems for the potential equation can be classified as fol-lows, g(r) a known function:

(a) (i) The interior Dirichlet problem

∇2u(r) = f(r), r ∈ Ω

u(r) = g(r), r ∈ ∂Ω.

(3.2)

(ii) The interior Neumann problem

∇2u(r) = f(r), r ∈ Ω

∂u(r)

∂n= g(r), r ∈ ∂Ω.

(3.3)

(b) (i) The exterior Dirichlet problem

∇2u(r) = f(r), r ∈ Ωc

u(r) = g(r), r ∈ ∂Ωc

|u(r)| = O(r−1) as r → ∞.

(3.4)

3.1 Mathematical background 23

(ii) The exterior Neumann problem

∇2u(r) = f(r), r ∈ Ωc

∂u(r)

∂n= g(r), r ∈ ∂Ωc

u(r) = O(r−1),∂u

∂r= O(r−2) as r → ∞.

(3.5)

(c) The mixed interior problem (Ω = Ωc for the exterior problem):

∇2u(r) = f(r), r ∈ Ω,

u(r) = g1(r), r ∈ ∂Ω1

∂u(r)

∂n= g2(r), r ∈ ∂Ω2

(3.6)

where ∂Ω1 ∪ ∂Ω2 = ∂Ω and ∂Ω1 ∩ ∂Ω2 = ∅.

(d) The Robin problem

∇2u(r) = f(r), r ∈ Ω

u(r) + k(r)∂u(r)

∂n= 0, r ∈ ∂Ω.

(3.7)

where k(r) is a known function defined on the boundary.

The Dirichlet problem (3.2) has a unique solution and the interior Neumannproblem (3.3) has a unique solution, up to the addition of an arbitrary constant,provided

∫

∂Ω

g(r) dS = 0. (3.8)

3.1.1 Green’s Theorems

Let u and v be two scalar functions defined in Ω which are continuous and admitcontinuous partial derivatives. For a fixed point r and a reference point z, the


function v(z; r) gives the potential in z due to a unit source in r. We have:

∇z · (u(z)∇zv(z; r)) = u(z)∇2zv(z; r) + ∇zu(z) · ∇zv(z; r), (3.9)

∇z · (v(z; r)∇zu(z)) = v(z; r)∇2zu(z) + ∇zv(z; r) · ∇zu(z), r, z ∈ Ω. (3.10)

Subtracting equation (3.10) from (3.9) and integrating gives∫

Ω

[∇z · (u(z)∇zv(z; r)) − ∇z · (v(z; r)∇zu(z))] dV(z)

=

∫

Ω

[u(z)∇2zv(z; r) − v(z; r)∇2

zu(z)] dV(z). (3.11)

If we apply Gauss’ Theorem on the left hand side of (3.11) we obtain:∫

Ω

[∇z · (u(z)∇zv(z; r)) − ∇z · (v(z; r)∇zu(z))] dV(z)

=

∫

∂Ω

(u(r ′)∇r ′v(r′; r) − v(r ′; r)∇r ′u(r ′)) · n(r ′) dS(r ′)

=

∫

∂Ω

(

u(r ′)∂v

∂n(r ′; r) − v(r ′; r)

∂u

∂n(r ′)

)

dS(r ′), (3.12)

where r ∈ Ω, n(r ′) is the outward unit normal at r ′ on the boundary ∂Ω and∂v/∂n = ∇v · n is the outward normal derivative of v in the direction of n.Substituting (3.12) into equation (3.11) yields Green’s second identity:

∫

Ω

[u(z)∇2zv(z; r) − v(z; r)∇2

zu(z)] dV(z)

=

∫

∂Ω

(

u(r ′)∂v

∂n(r ′; r) − v(r ′; r)

∂u

∂n(r ′)

)

dS(r ′), r ∈ Ω. (3.13)

In the BEM v is taken as the fundamental solution of the problem to be solved.We will derive an expression for the fundamental solution of the Laplace equa-tion in the next section.

3.1.2 Fundamental solution of the Laplace equation

The fundamental solution of the Laplace equation is the solution of the singu-larly forced Laplace equation

∇2zv(z; r) + δ(z; r) = 0, r, z ∈ Ω∞ , (3.14)


where z is the variable field point, r is the fixed location of the singular pointor pole and Ω∞ denotes the infinite domain which is the whole space in 3Dand the whole plane in 2D. Thus v(z; r) is the free space Green’s function. AGreen’s function of the first kind takes on a reference value of zero over theboundary ∂Ω whereas a Green’s function of the second kind, also called theNeumann function, has the property that its normal derivative vanishes over∂Ω [Pozrikidis, 2002, page 94]. The difference between the fundamental solu-tion and the Green’s function is therefore that the former is the general solutionto (3.14), while the latter is a particular solution which satisfies a set of speci-fied homogeneous boundary conditions. Thus the fundamental solution can beviewed as a generalised Green’s function also known as the free space Green’sfunction. The function δ(z; r) is the Dirac delta distribution which satisfies thefollowing properties [Pozrikidis, 2002, page 93]:

δ(z; r) =

0, r 6= z,

∞, r = z,

(3.15a)

∫

Ω

δ(z; r) dV(z) = 1, if r ∈ Ω, (3.15b)

∫

Ω

f(z)δ(z; r) dV(z) =

f(r), if r ∈ Ω,

0, if r /∈ Ω.

(3.15c)

Thus δ(z; r) plays the role of a unit source placed at the point r. In two dimen-sions we introduce polar coordinates such that u = u(r, θ) and in three dimen-sions we introduce spherical coordinates such that u = u(r, θ, ϕ). The solutionof (3.14) is symmetric with respect to the source point so that it is independent ofthe polar angle θ in two dimensions and independent of the azimuthal angle θ

in three dimensions. Thus it depends only on the variable r = ||r − z||. Applyingthe operator ∇2 to v(r) takes the form:

∇2v(r) =

(

d2

dr2+

α

r

d

dr

)

v(r) where α =

2 in 3D,

1 in 2D.


Solving the above equation we arrive at the fundamental solution as

v(r) =

K

rin 3D,

K ln 1r in 2D.

(3.16)

Due to the singular character of the solutions the functions v(r) reach an infinitevalue when r → 0, that is, when z tends to the fixed point r. To find the valuesof the constant K, we apply the properties of the Dirac delta distribution. Takea simple domain Ωε around the point r. Integrating (3.14) and applying theproperties (3.15) we obtain

∫

Ωε

∇2zv(z; r) dV(z) = −

∫

Ωε

δ(z; r) dV(z) = −1, r, z ∈ Ωε.

Applying the divergence theorem to the first integral yields

−1 =

∫

∂Ωε

∇r ′v(r′; r) · n(r ′) dS(r ′) =

∫

∂Ωε

∂v

∂n(r ′; r)dS(r ′). (3.17)

In 3D the domain Ωε is a sphere of radius ε whereas in 2D it is a circular discof radius ε. Due to the symmetric nature of the problem, ∂v/∂n = ∂v/∂r. Sincev is a function of r and ∂v/∂r is a constant on ∂Ωε defined by r = ε, expression(3.17) gives for both cases:

−1 =

∫

∂Ωε

∂v

∂rdS =

∂v

∂r

∣

∣

∣

∣

r=ε

· Sε,

in which Sε is the area, in 3D, and the length, in 2D, of ∂Ωε. In 3D:

∂v

∂r

∣

∣

∣

∣

r=ε

= −K

ε2, Sε = 4πε2.

Thus:

−1 = −K

ε2(4πε2) ⇒ K =

1

4π.

Similarly, in 2D:

∂v

∂r

∣

∣

∣

∣

r=ε

= −K

ε; Sε = 2πε.

Thus:

−1 = −K

ε(2πε) ⇒ K =

1

2π.


Hence the final expression for the fundamental solution is

v(z; r) =

1

4π

1

||z − r||in 3D,

1

2πln

1

||z − r||in 2D.

3.1.3 Integral properties of Green’s functions

Consider a singly or multiply connected control volume Ω bounded by a closedsurface or a collection of closed surfaces denoted by ∂Ω. Assume that all sur-faces are smooth. Pozrikidis in [Pozrikidis, 2002, page 97] has argued that inte-grating (3.14) over the control volume Ω and using the divergence theorem andthe distinctive properties of the delta distribution in three dimensions, we findthat Green’s functions satisfy the integral identity

∫

∂Ω

∂v

∂n(z; r) dS(z) =

1, r ∈ Ω,

1

2, r ∈ ∂Ω,

0, r ∈ Ωc,

(3.18)

where the unit normal vector n points into Ω as shown in Figure 3.2.

Ω n

∂Ω

Ωc

Figure 3.2: Definition of terms in equa-tion (3.18).


3.2 Integral formulation of the Laplace equation

The potential problem described in Chapter 2 is a boundary value problem forthe Laplace equation. We shall now discuss the integral formulation of bound-ary value problems.

3.2.1 Basic integral equation: Internal points

Here an internal point refers to a point inside the domain in which a solution tothe potential problem is sought. We shall now derive the integral representationof the potential at an internal point. Let r ∈ Ω. We start from Green’s secondidentity (3.13):

∫

Ω

(u(z)∇2zv(z; r) − v(z; r)∇2

zu(z))dV(z)

=

∫

∂Ω

(

u(r ′)∂v

∂n(r ′; r) − v(r ′; r)

∂u

∂n(r ′)

)

dS(r ′). (3.19)

The function u corresponds to the solution of the potential problem in the do-main Ω with the corresponding boundary conditions and the function v cor-responds to the fundamental solution of the Laplace equation. Because of thesingularity of the function v, the domain of integration where Green’s secondtheorem is applied must be defined isolating the point r. This is done by con-structing a ball Ωε of radius ε around it. In 3D Ωε equates to a sphere whereasit equates to a circular disc in 2D. Then the new domain of integration is nowΩ − Ωε with boundary ∂Ω + ∂Ωε, see Figure 3.3. Green’s second theorem isnow written as

limε→0

∫

Ω−Ωε

u(z)∇2zv(z; r) dV(z) − lim

ε→0

∫

Ω−Ωε

v(z; r)∇2u(z)dV(z)

= limε→0

∫

∂Ω

(

u(r ′)∂v

∂n(r ′; r) − v(r ′; r)

∂u

∂n(r ′)

)

dS(r ′)

+ limε→0

∫

∂Ωε

(

u(r ′)∂v

∂n(r ′; r) − v(r ′; r)

∂u

∂n(r ′)

)

dS(r ′) (3.20)

3.2 Integral formulation of the Laplace equation 29

ε

Ω − Ωε

∂Ωε

r

r ′

z

Ωε

∂Ω

n

n

Figure 3.3: Definition of the domain of integration forinternal points.

z ∈ Ω − Ωε, r ∈ Ωε.

Using (3.15), since the point r is outside the domain of integration, it implies thatthe first domain integral is zero. The second domain integral evaluates to

limε→0

∫

Ω−Ωε

v(z; r)∇2u(z)dV(z) = limε→0

∫

Ω−Ωε

v(z; r)f(z)dV(z)

=

∫

Ω

v(z; r)f(z)dV(z) = 0. (3.21)

Note that for the Laplace problem in which we are interested, f(z) = 0 so thatthis integral is straight away equal to zero.

The first part of the second integral of the right hand side of equation (3.20) istransformed as

limε→0

∫

∂Ωε

u(r ′)∂v

∂n(r ′; r)dS(r ′)

= limε→0

[∫

∂Ωε

(u(r ′) − u(r))∂v

∂n(r ′; r)dS(r ′) + u(r)

∫

∂Ωε

∂v

∂n(r ′; r)dS(r ′)

]

(3.22)

The limit of the first part of the right-hand side of equation (3.22) is zero foru Holder continuous [Parıs and Canas, 1997]. Now consider the second term


of the right-hand side of (3.22). In 2D, the boundary ∂Ωε is a circumference ofradius ε and

dS = εdθ,∂v

∂n=

1

2πε.

Then:

limε→0

u(r)

∫

∂Ωε

∂v

∂n(r ′; r)dS(r ′) = lim

ε→0u(r)

∫2π

0

1

2πεεdθ = u(r). (3.23)

In 3D we use spherical coordinates such that

dS = ε2 cos ϕ dϕ dθ,∂v

∂n=

1

4πε2.

Then:

limε→0

u(r)

∫

∂Ωε

∂v

∂n(r ′; r)dS(r ′) = lim

ε→0u(r)

∫2π

0

(∫π/2

−π/2

1

4πε2ε2 cos ϕ dϕ

)

dθ

=u(r)

4π

∫2π

0

(∫π/2

−π/2

cos ϕ dϕ

)

dθ = u(r). (3.24)

Now taking the second term of the second integral of the right-hand side of(3.20), for the 2D case:

limε→0

∫

∂Ωε

v(r ′; r)∂u

∂ndS(r ′) = lim

ε→0

(

−ln ε

2π

∫

∂Ωε

∂u

∂n(r ′) dS(r ′)

)

= 0 (3.25)

due to the fact that the integral of the flux for the Laplace problem through aclosed surface is zero. Similar reasoning for 3D yields

limε→0

∫

Ωε

v(r ′; r)∂u

∂ndS(r ′) = lim

ε→0

(

1

4πε

∫

∂Ωε

∂u

∂n(r ′) dS(r ′)

)

= 0. (3.26)

Therefore if we substitute equations (3.21), (3.24) and (3.26) in (3.20) and use(3.15) we get

u(r) =

∫

∂Ω

[

v(r ′; r)∂u

∂n(r ′) − u(r ′)

∂v

∂n(r ′; r)

]

dS(r ′), r ∈ Ω. (3.27)


The integral representation (3.27) shows that the potential u can be calculated atany point r inside the domain if its value u and the value of its derivative ∂u/∂n

are known on the boundary. The first and second integrals in (3.27) are calledthe single layer and double layer potentials respectively. In Section 3.2.4 we shalldiscuss the physical interpretation of these integrals.

3.2.2 Basic integral equation: Boundary points

The scheme followed in the definition of equation (3.20) starting from Green’ssecond theorem is also valid for a point r placed on the boundary ∂Ω althoughsome modifications are required. Let Ωi

ǫ and ∂Ωiǫ denote the parts of Ωǫ and

∂Ωǫ that belong to the domain Ω, see Figure 3.4.

ε

Ω

Ωiε

∂Ωiε

r

r ′

z

∂Ωε

∂Ω − ∂Ωεn

Figure 3.4: Definition of the domain of integration forboundary points.

Since we restrict ourselves to the study of the Laplace equation, the left-handside of equation (3.13) cancels and the right-hand side becomes

0 = limε→0

∫

∂Ω−∂Ωε

[

u(r ′)∂v

∂n(r ′; r) − v(r ′; r)

∂u

∂n(r ′)

]

dS(r ′)

+ limε→0

∫

∂Ωiε

[

u(r ′)∂v

∂n(r ′; r) − v(r ′; r)

∂u

∂n(r ′)

]

dS(r ′), r ∈ ∂Ωε. (3.28)


As was done in the previous section, the first term of the second integral in (3.28)can be transformed as (3.22) and then we only need to evaluate the ensuingsecond term. In 2D, this yields

limε→0

u(r)

∫

∂Ωiε

∂v

∂n(r ′; r)dS(r ′) = C(r)u(r), C(r) =

α(r)

2π,

where α(r) is the internal angle at point r, see Figure 3.5; the calculation is pre-sented in [Parıs and Canas, 1997, p 44] and [Katsikadelis, 2002, p 30].

∂Ω

Ωrα(r)

Figure 3.5: Definition of the internal angleat a boundary point r.

Therefore the first term of the second integral evaluates to

limε→0

∫

∂Ωiε

u(r ′)∂v

∂n(r ′; r) dS(r ′) = C(r)u(r). (3.29)

The second term of the second integral of (3.28) gives, in 2D,

limε→0

∫

∂Ωiε

v(r ′; r)∂u

∂n(r ′) dS(r ′) = lim

ε→0

(

−ln ε

2π

∫

∂Ωiε

∂u

∂n(r ′) dS(r ′)

)

= 0. (3.30)

In 3D, let us consider the case where the point r belongs to a flat surface and we


obtain

limε→0

u(r)

∫

∂Ωiε

∂v

∂n(r ′; r) dS(r ′)

= limε→0

u(r)

∫2π

0

(∫0

−π/2

1

4πε2ε2 cos ϕ dϕ

)

dθ

= u(r)

∫2π

0

(∫0

−π/2

1

4πcos ϕ dϕ

)

dθ =1

2u(r). (3.31)

The evaluation of the last term of (3.28) gives zero as explained in [Parıs andCanas, 1997]. Substituting the above results into equation (3.28) gives

1

2u(r) + lim

ε→0

∫

∂Ω−∂Ωε

u(r ′)∂v

∂n(r ′; r) dS(r ′)

= limε→0

∫

∂Ω−∂Ωε

v(r ′; r)∂u

∂n(r ′) dS(r ′) r ∈ ∂Ωε. (3.32)

The two integrals that remain are to be evaluated in the sense of Cauchy Princi-pal Value (CPV) integrals. The second integral has a weaker removable singu-larity and consequently has meaning as an improper integral. Therefore we canwrite equation (3.32) in the form

1

2u(r) =

∫

∂Ω

v(r ′; r)∂u

∂n(r ′) dS(r ′) − −

∫

∂Ω

u(r ′)∂v

∂n(r ′; r) dS(r ′) (3.33)

where −∫

indicates that this integral is to be evaluated in the CPV sense.

From Green’s second identity (3.13), it implies that when the point r is locatedoutside the domain Ω then we have

0 =

∫

∂Ω

[

v(r ′; r)∂u

∂n(r ′) − u(r ′)

∂v

∂n(r ′; r)

]

dS(r ′), r ∈ Ωc. (3.34)

Combining equations (3.27), (3.33) and (3.34) we obtain the following generalintegral equation

C(r)u(r) =

∫

∂Ω

v(r ′; r)∂u

∂n(r ′) dS(r ′) −

∫

∂Ω

u(r ′)∂v

∂n(r ′; r) dS(r ′). (3.35)


The function

C(r) =

α(r)

2πin 2D,

α(r)

4πin 3D,

(3.36)

depends on the internal angle α(r) at r described in Figure 3.5. All together, wesummarise the interior potential problem in the frame below:

For a flat surface, α(r) is π in 2D and 2π in 3D and we obtain the follow-ing equation for the interior problem:

C(r)u(r) =

∫

∂Ω

v(r ′; r)∂u

∂n(r ′) dS(r ′)−

∫

∂Ω

u(r ′)∂v

∂n(r ′; r) dS(r ′), (3.37)

where

C(r) =

1 r ∈ Ω,

1/2 r ∈ ∂Ω,

0 r ∈ Ωc,

(3.38)

and the normal n(r ′), Ω and Ωc are as shown here below

Ω

n(r ′)

∂Ω

Ωc

r ′


If we would like to find the solution of the Neumann problem, then the firstterm in the righthand side of (3.35) is known and the resulting equation is calleda Fredholm integral equation of the second kind. If we would like to find the so-lution of the Dirichlet problem, then the second term is known and the resultingequation is called a Fredholm integral equation of the first kind.

3.2.3 Infinite regions

In the case of the exterior potential problem that we shall be considering, thedomain considered is usually an infinite region. The boundary integral equa-tion (3.35) and all concepts presented so far are still valid for such a regionbut the functions involved must satisfy certain regularity conditions at infin-ity [C. A. Brebbia and Wrobel, 1984]. To derive these conditions we enclose thesurface ∂Ω with another surface ∂Ω of radius r as shown in Figure 3.6. In

Ω

∂Ω

∂Ω

Ωc

Figure 3.6: Infinite regions.

two dimensions ∂Ω is a circle; in three dimensions it is a sphere. For the finitedomain enclosed between ∂Ω and ∂Ω equation (3.35) implies that

C(r)u(r) =

∫

∂Ω

v(r ′; r)∂u

∂n(r ′) dS(r ′) +

∫

∂Ω

v(r ′; r)∂u

∂n(r ′) dS(r ′)

−

∫

∂Ω

u(r ′)∂v

∂n(r ′; r) dS(r ′) −

∫

∂Ω

u(r ′)∂v

∂n(r ′; r) dS(r ′). (3.39)


If we let the radius r → ∞, then (3.39) will be valid for points in ∂Ω only if

limr→∞

∫

∂Ω

[

v(r ′; r)∂u

∂n(r ′) − u(r ′)

∂v

∂n(r ′; r)

]

dS(r ′) = 0. (3.40)

For three dimensional problems

v(r ′; r) = O(r−1),

∂v

∂n= O(r−2),

and

dS(r ′) = O(r2).

Therefore condition (3.40) is satisfied if

u(r ′) = O(r−1) as r → ∞, (3.41)

∂u

∂n= O(r−2) as r → ∞. (3.42)

These are the extra regularity conditions at infinity.

If condition (3.40) is satisfied, then we obtain the following equation for pointson the internal surface ∂Ω

Ce(r)u(r) =

∫

∂Ω

v(r ′; r)∂u

∂n(r ′) dS(r ′) −

∫

∂Ω

u(r ′)∂v

∂n(r ′; r) dS(r ′), (3.43)

where the function Ce(r) depends on the internal angle at r in the exterior prob-lem domain and is given by

Ce(r) =

4π − α(r)

4πin 3D

2π − α(r)

2πin 2D

(3.44)

with α(r) the internal angle corresponding to the interior problem domain at r.Thus for a smooth boundary we obtain


Ce(r) = C(r) =

1 r ∈ Ω,

1/2 r ∈ ∂Ω,

0 r ∈ Ωc.

(3.45)

Note that in equation (3.43) we still have the normal as given for the interiorproblem, that is, pointing into the domain. For the exterior problem, we shouldalso use the outward normal vector. Multiplying by −1 the terms that involvethe normal allows us to keep the same normal vector for the exterior problem asfor the interior problem. We summarise the result in the frame below:

The exterior potential problem is given by the equation:

C(r)u(r) =

∫

∂Ω

u(r ′)∂v

∂n(r ′; r) dS(r ′)−

∫

∂Ω

v(r ′; r)∂u

∂n(r ′) dS(r ′), (3.46)

where

C(r) =

1 r ∈ Ω,

1/2 r ∈ ∂Ω,

0 r ∈ Ωc,

(3.47)

and the normal n(r ′), Ω and Ωc are as shown here below.

Ω

n(r ′)

∂Ω

Ωc

r ′


An integral equation for an exterior and interior Dirichlet problem using a

single layer potential

Let ue ∈ C2(Ωc) ∩ C1(Ωc) and ui ∈ C2(Ω) ∩ C1(Ω) denote the functions in the

external and the internal problems, see Figure 3.7:

∇2ue(r) = 0, r ∈ Ωc

ue(r) = g(r), r ∈ ∂Ωc,

(3.48)

∇2ui(r) = 0, r ∈ Ω

ui(r) = g(r), r ∈ ∂Ω.

(3.49)

Also let ue satisfy the extra condition in (3.4). Then, for ue, we have

Ω Ωc

n

ui

ue

Figure 3.7: The exterior and interiorproblem domains in equations (3.48) and(3.49).

∫

∂Ω

v(r ′; r)∂ue

∂n(r ′) dS(r ′) −

∫

∂Ω

ue(r ′)∂v

∂n(r ′; r) dS(r ′)

=

0 r ∈ Ω

−ue(r) r ∈ Ωc

−12ue(r) r ∈ ∂Ω.

(3.50)

And for ui we have∫

∂Ω

v(r ′; r)∂ui

∂n(r ′) dS(r ′) −

∫

∂Ω

ui(r ′)∂v

∂n(r ′; r) dS(r ′)

=

0 r ∈ Ωc

ui(r) r ∈ Ω12ui(r) r ∈ ∂Ω.

(3.51)


Define, for r ∈ ∂Ω

ui(r) = limx→r

ui(x), x ∈ Ω; ue(r) = limx→r

ue(x), x ∈ Ωc;

[u(r)] = ui(r) − ue(r), r ∈ ∂Ω.

Similarly, define

∂ui(r)

∂n= lim

x→rn(r) · ∇ui(x), x ∈ Ω;

∂ue(r)

∂n= lim

x→rn(r) · ∇ue(x), x ∈ Ωc;

[

∂u(r)

∂n

]

=∂ui(r)

∂n−

∂ue(r)

∂n, r ∈ ∂Ω.

Subtracting (3.51) from (3.50) we obtain

∫

∂Ω

[u(r)]∂v

∂n(r ′; r) dS(r ′) −

∫

∂Ω

v(r ′; r)

[

∂u(r)

∂n

]

dS(r ′)

=

−ui(r) r ∈ Ω

−ue(r) r ∈ Ωc

−12(ui(r) + ue(r)) r ∈ ∂Ω.

(3.52)

Now we have, since [u] = 0 on ∂Ω,

u(r) =

∫

∂Ω

[

∂u

∂n

]

v(r ′; r) dS(r ′), r /∈ ∂Ω, (3.53)

and since (ui + ue)/2 = g(r) on ∂Ω,

g(r) =

∫

∂Ω

[

∂u

∂n

]

v(r ′; r) dS(r ′), r ∈ ∂Ω.

Thus, writing

q =

[

∂u

∂n

]

we are led to the following integral equation:

Given g, find q such that

1

4π

∫

∂Ω

q(r ′)

||r − r ′||dS(r ′) = g(r), r ∈ ∂Ω. (3.54)


This is a Fredholm integral equation of the first kind with weakly singular sym-metric kernel [Johnson, 1987]. It is a representation of the solution of the exteriorand interior Dirichlet problems using a single layer potential. After determiningq from (3.54) we obtain the solutions ue and ui of (3.48) and (3.49) respectivelyusing the formula (3.53).

An exterior Dirichlet problem with double layer potential

We will now show in this section how an exterior Dirichlet problem can be for-mulated with a double layer potential. This section and the following two aremainly based on [Johnson, 1987]. Consider the exterior Dirichlet problem

∇2u(r) = 0, r ∈ Ωc,

u(r) = u0, r ∈ ∂Ω.

(3.55)

Ω

∂Ω

Ωc n

Figure 3.8: Definition of terms in theDirichlet problem (3.55).

The solution of (3.55) can be extended to the interior of Ω by letting u satisfy

∇2u(r) = 0 r ∈ Ω,

u(r) = u0 r ∈ ∂Ω.

(3.56)

Let us replace the Dirichlet boundary condition in (3.56) by a Neumann bound-ary condition so that we have the following interior Neumann problem


∇2u(r) = 0 r ∈ Ω,

∂u(r)∂n

≡ ∂ui

∂n= g(r) r ∈ ∂Ω.

(3.57)

where

g =∂ue

∂n.

Thus, in this case

[

∂u

∂n

]

= 0 and (3.52) gives

ue(r) + ui(r)

2= −

∫

∂Ω

[u(r)]∂v

∂n(r ′; r) dS(r ′) r ∈ ∂Ω.

Therefore

ue(r) ≡ −ui(r) − ue(r)

2+

ue(r) + ui(r)

2

= −ϕ(r)

2−

∫

∂Ω

ϕ(r)∂v

∂n(r ′; r) dS(r ′) (3.58)

where ϕ = [u] = ui−ue. Since ue(r) = u0(r) for r ∈ ∂Ω, we obtain the followingFredholm integral equation of the second kind

ϕ(r)

2+

∫

∂Ω

ϕ(r)∂v

∂n(r ′; r) dS(r ′) = −u0(r), r ∈ ∂Ω, (3.59)

which is a representation of the exterior Dirichlet problem using a double layerpotential [Johnson, 1987, page 220]. From equation (3.52) we note that for r /∈

∂Ω,

u(r) = −

∫

∂Ω

ϕ(r)∂v

∂n(r ′; r) dS(r ′). (3.60)

Thus after solving (3.59) we obtain ϕ(r) which is then used in (3.60) to obtain asolution for ue(r) for r ∈ Ωc.


An exterior Neumann problem with single layer potential

Now consider the exterior Neumann problem (3.5). Associate with it the interiorDirichlet problem

∇2u = 0, in Ω,

u = ue on ∂Ω.

(3.61)

Then from (3.52), since [u] = 0, we have the following single layer representationfor the exterior Neumann problem:

u(r) =

∫

∂Ω

[

∂u(r ′)

∂n

]

v(r ′; r) dS(r ′). (3.62)

Also for r ∈ ∂Ω we have

1

2

∂ui

∂n+

∂ue

∂n

=

∫

∂Ω

[

∂u

∂n

]

∂v

∂n(r ′; r) dS(r ′),

then

∂ue

∂n≡ −

1

2

∂ui

∂n−

∂ue

∂n

+

1

2

∂ui

∂n+

∂ue

∂n

= −1

2q(r) +

∫

∂Ω

q(r ′)∂v

∂n(r ′; r) dS(r ′). (3.63)

where q =[

∂u∂n

]

. Using the Neumann boundary condition we get the integralequation

1

2q(r) −

∫

∂Ω

q(r ′)∂v

∂n(r; r ′) dS(r ′) = −g(r), r ∈ ∂Ω. (3.64)

This is a Fredholm equation of the second kind whose solution q(r) is used in(3.62) to obtain the solution for ue(r).

3.2.4 Physical interpretation of single and double layer potentials

Let y ∈ R3 be a point where a single unit charge has been placed. The electric

field at point x induced by the charge at point y is given by

E(x) =x − y

4π||x − y||3.


The associated potential function is

u(x) =1

4π||x − y||, i.e, E(x) = −∇xu.

For a charge distribution of density ρ in a domain Ω ⊆ R3, the potential at x ∈ R

3

associated with the electric field generated by these charges is

u(x) =

∫

Ω

ρ(y)

4π||x − y||dV(y).

Similarly, we can calculate the potential associated with the electric field gen-erated by a charge distribution on a surface ∂Ω. This layer potential is givenby

u(x) =

∫

∂Ω

ρ(y)

4π||x − y||dS(y).

We note that

u(x) =

∫

∂Ω

ρ(y)

4π||x − y||dS(y) =

∫

∂Ω

v(y − x)ρ(y) dS(y)

where v is the fundamental solution of Laplace’s equation in R3 presented in

Section 3.1.2 on page 24. Therefore the single layer potential

u(x) = −

∫

∂Ω

v(y − x)ρ(y) dS(y)

is a multiple of the potential induced by a charge distribution of density ρ on asurface ∂Ω.

Consider

¯u =

∫

∂Ω

∂v

∂n(y)(x − y)h(y) dS(y).

where h(y) is a continuous function on S. Fix some t > 0. Suppose we havea charge distribution on a surface S in R

3 such that the charge density at anypoint is given by t−1ρ(y). In addition, suppose we have a charge distribution ofopposite sign on a parallel surface

St = y + t n(y) : y ∈ S,

such that the charge density on St is given by −t−1ρ(y). Then the electric fieldat any point x ∈ R

3 generated by these electric charges is given by

E(x) = −∇u(x)


where u is the associated potential given by

u(x) =

∫

S

[

1

||x − y||−

1

||x − (y + t n(y))||

]

ρ(y)

tdS(y).

As t → 0,

[

1

||x − y||−

1

||x − (y + tn(y))||

]

1

t→ −

∂

∂n

(

1

||x − y||

)

Therefore the double layer potential can be thought of as a multiple of the po-tential induced by a double layer of charges of opposite sign on ∂Ω. Thus theelectric field is discontinuous over a single layer and continuous over a doublelayer.

3.2.5 The electric field intensity

The electric field intensity E corresponding to the potential u is given by

E(r) = −∇ru = −∂u

∂xi −

∂u

∂yj −

∂u

∂zk. (3.65)

Therefore the electric field intensity at any point in Ω is obtained by direct dif-ferentiation of equation (3.46) with C(r) = 1. For the sake of clarity, let ∂v/∂n ′

denote n(r ′) · ∇r ′v, then we have

E(r) = −∇ru(r)

= −∇r

[∫

∂Ω

u(r ′)∂v

∂n ′(r ′; r) dS(r ′) −

∫

∂Ω

v(r ′; r)∂u

∂n ′(r ′) dS(r ′)

]

= −

∫

∂Ω

u(r ′)∇r

(

∂v

∂n ′(r ′; r)

)

dS(r ′) +

∫

∂Ω

∂u

∂n ′(r ′)∇rv(r

′; r)(r ′) dS(r ′).


Now,

v(r ′; r) =1

4π·

1

||r ′ − r||,

∂v

∂n ′(r ′; r) = n(r ′) · ∇r ′v(r

′; r)

= n(r ′) ·

∂

∂x ′,

∂

∂y ′,

∂

∂z ′

v(r ′; r).

∂v

∂x ′(r ′; r) =

1

4π·

∂

∂x ′

(x ′ − x)2 + (y ′ − y)2 + (z ′ − z)2

−1/2

=1

4π·−(x ′ − x)

||r ′ − r||3.

Similar results for ∂v/∂y ′ and ∂v/∂z ′ are obtained in the same way so that wehave

∇r ′v(r′; r) =

1

4π·

r − r ′

||r ′ − r||3. (3.66)

Therefore

∂v

∂n ′(r ′; r) = n(r ′) · ∇r ′v(r

′; r) =n(r ′) · (r − r ′)

4π||r ′ − r||3.

Then

∇r

(

∂v

∂n ′(r ′; r)

)

= ∇rn(r ′) · (r − r ′)

4π||r ′ − r||3.


For the x−component,

∂

∂x

n(r ′) · (r − r ′)

4π||r ′ − r||3=

∂

∂x

(x − x ′)nx′ + (y − y ′)ny′ + (z − z ′)nz′

||r ′ − r||3,

=||r ′ − r||3nx′ − [(r − r ′) · n(r ′)] · [−3(x ′ − x)||r ′ − r||]

||r ′ − r||6,

=nx′

||r ′ − r||3+

3(x ′ − x)(r − r ′) · n(r ′)

||r ′ − r||5.

The results for ∂/∂y and ∂/∂z can be obtained in a similar way so that

∇r

(

∂v

∂n ′(r ′; r)

)

=n(r ′)

||r ′ − r||3+

3(r ′ − r)[(r − r ′) · n(r ′)]

||r ′ − r||5.

Therefore

E(r) = −1

4π

∫

∂Ω

u(r ′)

[

n(r ′)

||r ′ − r||3+

3(r ′ − r)[(r − r ′) · n(r ′)]

||r ′ − r||5

]

dS(r ′)+

1

4π

∫

∂Ω

∂u

∂n(r ′)

(r ′ − r)

||r ′ − r||3dS(r ′). (3.67)

From equation (3.67), it implies that the x−component of the electric field inten-sity is given by

Ex = −1

4π

∫

∂Ω

u(r ′)

[

nx′

||r ′ − r||3+

3(x ′ − x)

||r ′ − r||5[(r − r ′) · n(r ′)]

]

dS(r ′)

+1

4π

∫

∂Ω

q(r ′)(x ′ − x)

||r ′ − r||3dS(r ′). (3.68)

Equations for the y and z components are similar.

We have to note that the differentiation with respect to n(r ′) is carried out atthe surface point r ′(x ′, y ′, z ′) while the differentiation with respect to x, y or z iscarried out at the field point r(x, y, z).

3.3 The BEM for the Laplace equation 47

3.3 The BEM for the Laplace equation

The essence of the boundary element method is the discretisation of the bound-ary ∂Ω into small portions called boundary elements. On each of these ele-ments an assumption is made on the function u which results in different typesof elements; one may use constant elements, linear elements and parabolic orquadratic elements. On each of these elements we apply the boundary integralequation (3.46) which results in what is called the discretisation of the integralequation. The implementation of the BEM may be summarised in the followingsteps as we shall see in the subsequent sections:

1. discretisation of the integral equation,

2. evaluation of the integration constants,

3. application of the boundary conditions,

4. generation of the final system of equations,

5. calculation of the potential at internal points.

In this project we use constant elements in which the values of the function u

and its normal derivative ∂u/∂n are assumed constant over each element.

3.3.1 Constant elements

The boundary surface ∂Ω is divided into small surfaces ∂Ωj, j = 1, 2, . . . , N,

called elements. In this thesis we have used triangular elements and a pictureof such elements is shown in Figure 3.9. To generate the triangles we definestep lengths hx in the x-direction, hy in the y-direction and hz in the z-direction.For a cuboid boundary ∂Ω, we define a uniform mesh of rectangles at each sur-face, and divide each rectangle along the diagonal to form two triangles, seeFigure 3.10.

On each surface ∂Ωj we assume the functions u and ∂u/∂n to be constant withvalues u and q respectively. The constant values of the functions refer to the cen-tres of the elements which will be called the nodes. Thus we have N nodal pointson the boundary and at each node we have two unknowns u and q. Thereforewe have a total of 2N unknowns. For a unique solution, one of the unknowns


0 0.5 1 1.5 2 2.5 3 3.5 4x 0

24

68

10

y

0

1

2

3

4

5

6

z

Figure 3.9: A mesh generated with step length h = 1.0 inall directions, only three faces shown for clarity.

rj+1

rj

u = uj

u = uj+1

q = qj

q = qj+1

hx

hy

∂Ωj

∂Ωj+1

Figure 3.10: An example of two elements on a surface in the xy-plane.


must be prescribed at every nodal point and then the other N will be obtainedfrom the linear system yet to be derived.

Let uj, qj denote the values of u and q respectively on element ∂Ωj. For a givenpoint ri on ∂Ω the discretised version of the integral equation (3.46) on page 37is given by

1

2ui −

N∑

j=1

uj

∫

∂Ωj

∂v

∂n(r ′; ri) dS(r ′) = −

N∑

j=1

qj

∫

∂Ωj

v(r ′; ri) dS(r ′). (3.69)

The integrals in the above equation relate the node ri where the fundamentalsolution is applied to the nodes rj (j = 1, . . . , N). Their values give the contribu-tion of the nodal values uj and qj to the formation of the value ui. We denotethese influence coefficients by Hij and Gij respectively and define them as

Hij =

∫

∂Ωj

∂v(r ′j; ri)

∂ndS(r ′) , Gij =

∫

∂Ωj

v(r ′j; ri) dS(r ′). (3.70)

With the new notation, equation (3.69) becomes

−1

2ui +

N∑

j=1

Hijuj =

N∑

j=1

Gijqj. (3.71)

Setting

Hij = Hij −1

2δij, (3.72)

where δij is the kronecker delta, the final equation becomes

N∑

j=1

Hijuj =

N∑

j=1

Gijqj. (3.73)

The equation (3.73) is applied to all the collocation points ri (i = 1, . . . , N) yield-ing a system of N linear algebraic equations, which are arranged in matrix form

Hu = Gq, (3.74)

where H and G are N×N square matrices and u and q are vectors of dimensionN. For our problem (2.23) on page 19, the potential satisfies boundary conditionsat:

• Γ1, the anode surface;


• Γ2, the cathode surface and

• Γ3, the insulating hull surface.

For generality, let us assume a part ∂Ω1 of the boundary on which u is given anda part ∂Ω2 on which q is given. Let these two parts be discretised into N1 andN2 constant elements respectively such that N1 + N2 = N. Thus we still have N

unknowns, N−N1 values of u on ∂Ω2 and N−N2 values of q on ∂Ω1 which areto be determined from the system (3.74). Before we solve the system we needto separate the unknown from the known quantities. Therefore we partition thematrices H and G and write (3.74) as

(H1 H2)

u1

u2

= (G1 G2)

q1

q2

, (3.75)

where u1 and q2 denote the prescribed quantities on ∂Ω1 and ∂Ω2 respectively.We carry out the multiplications and move all the unknowns to the left handside of the equation so that we obtain

Ax = b (3.76)

where

A = (H2 − G1) , (3.77)

x =

u2

q1

,

b = −H1u1 + G2q2.

If the N1 points and the N2 points where the values of u and respectively q areprescribed are not consecutive then the partitioning of the matrices in (3.75) ispreceded by an appropriate rearrangement of columns in H and G. By solving(3.76) we obtain the unknown boundary quantities u and q. Therefore we nowhave all the boundary quantities. We can then compute the solution u(r) at anypoint r ∈ Ω from equation (3.46) with C(r) = 1. We use the same discretisationas in (3.69) to obtain

u(ri).=

N∑

j=1

Hijuj −

N∑

j=1

Gijqj. (3.78)

The coefficients Hij and Gij are again computed from (3.70) but the boundarypoint ri is replaced by the field point ri ∈ Ω.


The electric field intensity

Discretising equation (3.67) we get the following expression in which, the coef-ficients Hij and Gij are indicated as vectors since they contain the terms for thex, y and z components of the electric field intensity:

E(ri) = −

N∑

j=1

Hijuj +

N∑

j=1

Gijqj, (3.79)

where

Hij =1

4π

∫

∂Ωj

[

n(r ′j)

||r ′ − r||3+

3(r ′j − ri)

||r ′ − r||5[(ri − r ′j) · n(r ′)]

]

dS(r ′), (3.80)

and

Gij =1

4π

∫

∂Ωj

r ′ − r

||r ′ − r||3dS(r ′). (3.81)

3.3.2 Calculation of the matrix elements

From the above discussion we see that we have to evaluate the integrals in-volved in order to complete our numerical formulation of the problem. In thesequel we show how to evaluate these integrals using numerical methods. Let(∂Ωj) = ∂Ω1, ∂Ω2, . . . ∂ΩN be a triangulation of ∂Ω. Then

Gij =

∫

∂Ωj

v(ri; r′

j) dS(r ′j). (3.82)

The simplest way for calculating the integrals above is to use Gauss quadra-ture rules for triangles directly. First we need to map the triangle, say k, in thephysical 3D space to a right isosceles triangle in the ξ1ξ2-parametric plane withcoordinates (0,0), (0,1) and (1,0). To this end, for a given node on the referencetriangle the following transformation gives the corresponding node on the realsurface element

r = ξ1r1 + ξ2r2 + ξ3r3 (3.83)


where ξ3 = 1 − ξ1 − ξ2 and rl = (xl, yl, zl); l = 1, 2, 3 are the coordinates ofthe vertices of the triangle k. Thus we have the two tangential vectors in thedirection of ξ1 and ξ2 axes over the triangle as

T1 =∂r

ξ1

= r1 − r3 and T2 =∂r

∂ξ2

= r2 − r3. (3.84)

Then the integral of a function f(r) over k in physical space is given by

∫

k

f(r) dS = |T1 × T2|

∫1

0

∫1−ξ1

0

f(r(ξ1, ξ2)) dξ2 dξ1 (3.85)

If the function f(r) is nonsingular over the surface of the triangle then the inte-gral (3.85) may be computed accurately using numerical quadrature rules for atriangle according to the formula [Pozrikidis, 2002, page 119] and [Katsikadelis,2002, page 302]:

∫

k

f(r) dS.= Ak

NG∑

k=1

f(r(ξk1, ξk

2))wk, (3.86)

where;

NG is the number of quadrature points,

(ξk1, ξk

2) are the coordinates of the kth quadrature point,

wk is the weight corresponding to the kth quadrature point, and

Ak = 12|T1 × T2| is the area of the triangle.

Therefore if we apply the formula (3.86) to the integral (3.82) on a triangle wherev(ri; rj) is nonsingular we obtain

Gij.= Aj

NG∑

k=1

v(ξk1, ξk

2, ξk3)wk, (3.87)

where Aj is the area of ∂Ωj. Recall that

v(r ′j; ri) =1

4π||ri − r ′j||.


Let mxi, myi, mzi denote the coordinates of the i-th point ri, which is the cen-troid of the i-th triangle, then

v(x ′

j, y′

j, z′

j) =1

4π√

(mxi − x ′

j)2 + (myi − y ′

j)2 + (mzi − z ′

j)2, (3.88)

where r ′j = (x ′

j, y′

j, z′

j) is a point on the j-th integration triangle. Here we haveused v(r ′j; ri) = v(r ′j) for simplicity of notation. Using the transformation in(3.83), we have

x ′

j = xj1ξ

k1 + x

j2ξ

k2 + x

j3ξ

k3

y ′

j = yj1ξ

k1 + y

j2ξ

k2 + y

j3ξ

k3 (3.89)

z ′

j = zj1ξ

k1 + z

j2ξ

k2 + z

j3ξ

k3

where (xji, y

ji, z

ji), i = 1, 2, 3 are the coordinates of the vertices of the j-th integra-

tion triangle. Thus

v(ξk1, ξk

2, ξk3) = 1/(4π(mxi − (x

j1ξ

k1 + x

j2ξ

k2 + x

j3ξ

k3))2+

(myi − (yj1ξ

k1 + y

j2ξ

k2 + y

j3ξ

k3))2+

(mzi − (zj1ξ

k1 + z

j2ξ

k2 + z

j3ξ

k3))21/2) (3.90)

The points (ξk1, ξk

2, ξk3) with their corresponding weights wk can be found in var-

ious mathematical handbooks. For example, in [Katsikadelis, 2002, page 303]there is such a table for integration rules which are accurate for polynomials oforder one to five, in [Johnson, 1987, page 246] similar weights are given for or-der one to three and in [Bonnet, 1995, page 358] a table of Gauss points and thecorresponding weights for integration over the standard reference triangle aregiven for values of NG = 3, 6, 7, 16, 19, see Appendix A.

To calculate the integrals corresponding to Hij in (3.70) we have to evaluate thenormal derivative of v(r; r ′) at the point r ′. To this end we proceed as follows:

∂v(r ′; r)

∂n= ∇r ′v · n(r ′) =

∂v

∂r∇r ′r · n(r ′), n(r ′) =

T1 × T2

|T1 × T2|.

Now,

r = ||r − r ′|| = (x − x ′)2 + (y − y ′)2 + (z − z ′)21/2.


Then,

∂r

∂x ′=

−2(x − x ′)

2(x − x ′)2 + (y − y ′)2 + (z − z ′)21/2= −

x − x ′

r.

Similarly

∂r

∂y ′= −

y − y ′

rand

∂r

∂z ′= −

z − z ′

r.

Also,

∂v

∂r= −

1

4πr2.

Therefore,

∂v

∂n= −

1

4πr2

(

−x − x ′

r, −

y − y ′

r, −

z − z ′

r

)

· n(r ′)

=1

4πr3

(

(x − x ′)nx′ + (y − y ′)ny′ + (z − z ′)nz′

)

where n(r ′) = (nx′ , ny′ , nz′).

So from (3.70)

Hij =

∫

∂Ωj

∂v(r ′; ri)

∂ndS(r ′j) (3.91)

=1

4π

∫

∂Ωj

1

r3

(

(xi − x ′)nx′ + (yi − y ′)ny′ + (zi − z ′)nz′

)

dS(r ′). (3.92)

For a triangle on which the integrand in (3.92) is non-singular, the individualintegrals can be evaluated using Gaussian quadrature in a manner similar tothat in (3.87) which yields

Hij.=

Aj

4π

NG∑

k=1

1

r3(ξk1, ξk

2, ξk3)

(

(xi − x ′

j(ξk1, ξk

2, ξk3))nx′

j+

(yi − y ′

j(ξk1, ξk

2, ξk3))ny′

j+ (zi − z ′

j(ξk1, ξk

2, ξk3)−)nz′

j

)

wk, (3.93)


r3(ξk1, ξk

2, ξk3) = ((x ′

j(ξk1, ξk

2, ξk3) − xi)

2+

(y ′

j(ξk1, ξk

2, ξk3) − yi)

2 + (z ′

j(ξk1, ξk

2, ξk3) − zi)

2)3/2. (3.94)

Next we consider the case when the integrands in (3.82) and (3.92) are singular.This occurs when the collocation point belongs to the element over which weare integrating. For the case of (3.82), we remove the singularity by subdividingthe triangle into three sub-triangles and introducing polar coordinates with thepoint of singularity as the origin as shown in Figure 3.11. The integrals are thenevaluated analytically in the from

Gii =

θ1∫

0

R1(θ)∫

0

dRdθ +

θ1+θ2∫

θ1

R2(θ)∫

0

dRdθ +

2π∫

θ1+θ2

R3(θ)∫

0

dRdθ. (3.95)

The result of (3.95) is given in [C. A. Brebbia and Wrobel, 1984, page 93] and has

1 2

3

R1(θ)θ1

R2(θ)

θ2

R3(θ)

θ3

α1

α2

α3

L99 −−−−−−−−−− 99K

L99−−−−−−−−−−−−

99K

L99−−−−−−−−−−−−

99K

r12

r13 r23

Figure 3.11: Definition of the terms used in equation (3.96).


also been used in [Ramachandran, 1994, page 241] as follows:

Gii =2Ai

3

1

r23

ln

(

tan[(θ1 + α2)/2]

tan(α2/2)

)

+1

r31

ln

(

tan[(θ2 + α3)/2]

tan(α3/2)

)

+

1

r12

ln

(

tan[(θ3 + α1)/2]

tan(α1/2)

). (3.96)

The geometrical terms in this integral are shown in Figure 3.11.

For the case of (3.91), singularity occurs when the collocation point belongs tothe element over which we are integrating. In this case the vector ri−r ′j lies in thesurface of the element perpendicular to the normal n(r ′). As a result ∂v/∂n = 0

and hence, from (3.72),

Hii = 0, ⇒ Hii = −1

2. (3.97)

Hence, using (3.87), (3.93), (3.96) and (3.97), we are able to assemble the matricesH and G and consequently the matrix A in (3.76) through the procedures in(3.75) and (3.77). The matrix A is fully populated and nonsymmetric.

The integration for the layer potentials could also be done analytically thoughthe exercise is more cumbersome. For example, in [Rao et al., 1979] an analyticalsolution for the single layer potential is given, in [Rao et al., 1984] similar inte-grals are discussed and more references given where the analytic solutions canbe found. In [Eibert and Hansen, 1995, Graglia, 1993] these integrals have beenevaluated in relation to the method of moments.

In summary, we have discussed the boundary element formulation of Laplace’sequation. In Sections 3.1 and 3.2, the integral formulation of Laplace’s equationis presented. The calculation of the electric field intensity is also discussed. InSection 3.3.2 we have discussed the numerical implementation of the boundaryelements method for the Laplacian. Methods to calculate the single and doublelayer potentials have been discussed, analytical methods for singular integralsand Gaussian quadrature for nonsingular integrals. In Chapter 4, we shall runsimulations of this numerical formulation and discuss our results for the modelproblem discussed in Section 2.5.

Chapter 4

Numerical results and discussion

In this chapter we discuss the results obtained from applying the numerical for-mulation outlined in Chapter 3 to the potential problem introduced in Chapter2. For the model problem explained in Section 2.5, we consider the domain Ω tobe, see Figure 4.1:

Ω =

(x, y, z) ∈ R3\Ωc

,

where

Ωc =

(x, y, z) ∈ R3 | 0 < x < 4, 0 < y < 10, 0 < z < 6

.

The anode surface is defined by

Γ1 = (x, y, z) ∈ R3 | x = 4, 2 ≤ y ≤ 4, 2 ≤ z ≤ 4,

the cathode surface by

Γ2 = (x, y, z) ∈ R3 | 1 ≤ x ≤ 3, y = 10, 2 ≤ z ≤ 4,

and the insulating surface by

Γ3 = ∂Ω\(Γ1 ∪ Γ2).

58 Chapter 4. Numerical results and discussion

z

y

x

10

6

4Ω

Ωc

AnodeCath

ode

Insulating hull

Figure 4.1: The domain for the potential problem (2.23).

For all the results presented in this chapter uniform triangular elements wereused everywhere for the discretisation of the boundary. In fact, in reference toFigure 3.10, we have used hx = hy = hz = h and generated results for differentvalues of h.

4.1 An example

For the model problem described in Section 2.5, we verified the formulation inChapter 3 using the function

f = 1/r, r =

√

(x − 2)2 + (y − 5)2 + (z − 3)2, (4.1)

where the source point (2, 5, 3) is located at the centre of the box. For the Dirich-let problem, analytically computed values of f were prescribed at the entire boxsurface. The BEM was then used to compute the values of ∂f/∂n at the surfaceand subsequently the values of f at some points outside the box. For the Neu-mann problem, values of ∂f/∂n were prescribed and for the mixed boundaryvalue problem, values of f were prescribed at the electrode surfaces and ∂f/∂n

at the rest of the surface. The results are shown in Tables 4.1, 4.2, and 4.3 re-spectively for points in the plane z = 3. The intensity corresponding to such apotential problem was also calculated and the field lines, normalised, are shownin Figure 4.2.

4.1 An example 59

Exact BEM, N = 496 BEM, N = 1984 BEM, N = 7936

u u Relative u Relative u RelativePosition ×10−1

×10−1 error ×10−1 error ×10−1 error

(-4,5,3) 1.66667 1.6688 1.2531×10−3 1.6674 4.356×10−4 1.66691 1.447×10−4

(-3,5,3) 2.00000 2.0010 4.896×10−4 2.0005 2.329×10−4 2.00018 8.84×10−5

(-2,5,3) 2.50000 2.4979 8.300×10−4 2.4997 1.047×10−4 2.49999 3.1×10−6

(-1,5,3) 3.33333 3.3213 3.6145×10−3 3.3310 7.079×10−4 3.33280 1.588×10−4

(5,5,3) 3.33333 3.3213 3.6145×10−3 3.3310 7.079×10−4 3.33280 1.588×10−6

(6,5,3) 2.50000 2.4979 8.300×10−4 2.4997 1.047×10−4 2.49999 3.1×10−6

(7,5,3) 2.00000 2.0010 4.896×10−4 2.0005 2.329×10−4 2.00018 8.81×10−5

(8,5,3) 1.66667 1.6688 1.2531×10−3 1.6673 4.356×10−4 1.66691 1.447×10−4

Table 4.1: Results for f = 1/r with Dirichlet boundary conditions, the sourcepoint is (2,5,3).

Exact BEM, N = 496 BEM, N = 1984

u u Relative u RelativePosition ×10−1 ×10−1 error ×10−1 error

(-4,5,3) 1.66667 1.66989 1.9338×10−3 1.66756 5.376×10−4

(-3,5,3) 2.00000 2.00191 9.564×10−4 2.00059 2.955×10−4

(-2,5,3) 2.50000 2.49768 9.291×10−4 2.49953 1.876×10−4

(-1,5,3) 3.33333 3.31607 5.1798×10−3 3.32919 1.2431×10−3

(5,5,3) 3.33333 3.31607 5.1798×10−3 3.32919 1.2431×10−3

(6,5,3) 2.50000 2.49768 9.291×10−4 2.49953 1.876×10−4

(7,5,3) 2.00000 2.00191 9.564×10−4 2.00059 2.955×10−4

(8,5,3) 1.66667 1.66989 1.9338×10−3 1.66756 5.376×10−4

Table 4.2: Results for f = 1/r with Neumann boundary conditions, the sourcepoint is (2,5,3).

From the tables we observe that the numerical solution of the Dirichlet problemis more accurate than the solution of the Neumann problem and that the solu-tion of the mixed boundary value problem is in between. The errors are partlyattributed to discretisation as we also see in the tables that the relative errors arereduced when we increase the number of elements used.


Exact BEM, N = 496 BEM, N = 1984

u u Relative u RelativePosition ×10−1 ×10−1 error ×10−1 error

(-4,5,3) 1.66667 1.66982 1.8930×10−3 1.66754 5.252×10−4

(-3,5,3) 2.00000 2.00184 9.208×10−4 2.00057 2.848×10−4

(-2,5,3) 2.50000 2.49760 9.586×10−4 2.49951 1.965×10−4

(-1,5,3) 3.33333 3.31599 5.2025×10−3 3.32917 1.2499×10−3

(5,5,3) 3.33333 3.31586 5.2426×10−3 3.32910 1.2700×10−3

(6,5,3) 2.50000 2.49728 1.0881×10−3 2.49938 2.491×10−4

(7,5,3) 2.00000 2.00151 7.571×10−4 2.00044 2.212×10−4

(8,5,3) 1.66667 1.66954 1.7247×10−3 1.66744 4.611×10−4

Table 4.3: Results for f = 1/r with mixed boundary conditions, the source pointis (2,5,3).

−5 0 5 10−5

0

5

10

15

x

y

Figure 4.2: Field lines corresponding to the potential f =

1/r in the plane z = 3.0.

4.2 Solution to the model problem 61

4.2 Solution to the model problem

Now, for our model problem, we prescribe the following linear boundary con-ditions:

u(r) = 0.5, r ∈ Γ1,

u(r) = −0.5, r ∈ Γ2,

∂u

∂n(r) = 0, r ∈ Γ3.

(4.2)

The potential around the box was calculated using BEM with different num-bers of elements. Tabel 4.4 shows the values of u computed along the line(y = 3, z = 3) which goes through the anode. As we have already seen inthe example, we expect the values for a higher number of elements to be moreaccurate. There is an indication that u(r) → 0 as r → ∞ which means thatour solution satisfies condition (3.41) required for the exterior problem. Thisbehaviour is also depicted in Figure 4.3.

Figures 4.4 to 4.6 show the potential and the electric field distribution in differentz-planes. We see again that the potential reduces when the distance from theelectrodes increases and that the current distribution varies as well.

In the plane z = 9.0 which is well above the box, we see that the field lines be-have as if there is a point source and sink above the anode and cathode respec-tively but displaced a little distance. This shows that in this plane, the currentflows in from below near the anode in an upward vertical orientation and re-turns to the cathode in a more vertical orientation. Hence there are points abovethe anode and above the cathode as shown in Figure 4.6 which shows a projec-tion of the electric field lines in the plane z = 9.

In Figure 4.7 we have scaled the electric field vectors for a clear view of thecurrent flow around the box in the plane z = 3.0. In this figure, the variationof the colours on the box is a representation of the variation of the potential onthe box surface. Thus red at the anode and blue at the cathode and the restin between. Due to symmetry of the solution the z-components of the electricfield vectors in the plane z = 3.0 are all negligible. However, in other planes,to compute the electric field vectors we have shown in this section, we haveused the x- and y- components to obtain a view of the field in the correspondingz-plane. That is, we have projections of the electric field lines in these planes.


u × 10−3

Position h = 2.00 h = 1.00 h = 0.500

(N = 124) (N = 496) (N = 1984)

(-4.0,3.0,3.0) -3.364629403 -2.389935812 -2.621423868(-3.5,3.0,3.0) -3.315538785 -2.304478296 -2.531666746(-3.0,3.0,3.0) -3.244141066 -2.196726509 -2.417979180(-2.5,3.0,3.0) -3.152069887 -2.070134995 -2.284460338(-2.0,3.0,3.0) -3.043032285 -1.931485440 -2.139262172(-1.5,3.0,3.0) -2.922453984 -1.791050453 -1.995080419(-1.0,3.0,3.0) -2.796487609 -1.661467466 -1.868319020(-0.5,3.0,3.0) -2.699916628 -1.554428285 -1.776048375(4.5,3.0,3.0) 335.6669754 344.1082863 347.7357228(5.0,3.0,3.0) 214.8988894 230.9242993 238.6588042(5.5,3.0,3.0) 149.1537113 162.8333565 170.0265585(6.0,3.0,3.0) 109.3441642 120.4420081 126.3556491(6.5,3.0,3.0) 83.50978899 92.51798577 97.27392723(7.0,3.0,3.0) 65.75284369 73.17127051 77.01154270(7.5,3.0,3.0) 52.99150975 59.20889734 62.34402711(8.0,3.0,3.0) 43.49888107 48.79864061 51.38952638

Table 4.4: Results for the model problem.


−0.3

−0.2

−0.1

0

0.1

0.2

0.3

−50

510

−50

510

15−0.4

−0.2

0

0.2

0.4

xy

u

N = 496

−0.3

−0.2

−0.1

0

0.1

0.2

0.3

−4 −2 0 2 4 6 8−4

−2

0

2

4

6

8

10

12

14

x

y

N = 496

−0.3

−0.2

−0.1

0

0.1

0.2

0.3

−50

510

−50

510

15−0.4

−0.2

0

0.2

0.4

xy

u

N = 1984

−0.3

−0.2

−0.1

0

0.1

0.2

0.3

−4 −2 0 2 4 6 8−4

−2

0

2

4

6

8

10

12

14

x

y

N = 1984

Figure 4.3: Surface (left) and pseudo colour (right) plots of the potential forh = 1.0 and 0.5 in the plane z = 3.0.


−0.3

−0.2

−0.1

0

0.1

0.2

0.3

−50

510

−50

510

15−0.4

−0.2

0

0.2

0.4

xy

u

−5 0 5 10−5

0

5

10

15

x

y

Figure 4.4: The potential (above) and field lines (below) in the planez = 3.0 for N=1984.


−0.15

−0.1

−0.05

0

0.05

0.1

0.15

−50

510

−50

510

15−0.2

−0.1

0

0.1

0.2

xy

u

−0.06

−0.04

−0.02

0

0.02

0.04

0.06

−50

510

−50

510

15−0.1

−0.05

0

0.05

0.1

xy

u

Figure 4.5: The potential in the planes z = 4.5 (above) and z = 6.0 (below).


−0.015

−0.01

−0.005

0

0.005

0.01

0.015

−50

510

−50

510

15−0.02

−0.01

0

0.01

0.02

xy

u

(a)

−5 0 5 10−5

0

5

10

15

x

y

(b)

Figure 4.6: The potential (a) and field lines (b) in the plane z = 9.0

which lies above the box.


Figure 4.7: Solution to the model problem. Electric fieldvectors are normalised. The colour variation denotes a vari-ation in the electric field strength for the arrows and the sizefor the electric potential on the box surface.


Figure 4.8 shows that the potential falls (rises) from the highest (lowest) valueat the anode (cathode) to zero as the distance from the box increases. This isan indication that the solution satisfies the condition u → 0 at infinity whichis required for the external problem. The closeness of the curves also indicatesthat, at distances away from the ships’ surface, the potential can well be approx-imated using either 124, 496 or 1984 elements. The solutions for N = 496 andN = 1984 are even closer to the exact solution.

Figure 4.9 is a plot of u outside the box calculated using different numbers ofelements but with the same number of quadrature points. The exact value of thepotential at x = 4.0 is u = 0.5. In the figure we calculate the potential very closeto x = 4.0 which is actually the location of the surface. We see that we can not usethe quadrature rule for points that are very close to the box surface. In regionswithin a distance of about 0.05 the quadrature rule fails and underestimates thepotential. However a better approximation is obtained by increasing the numberof elements.

In Figure 4.10 we see that for very small distances r, say less than 0.1, it is betterto use more quadrature points but this does not matter for larger distances.

4.3 Current flow out of the box

In this section we investigate the amount of current that flows out of the elec-trodes.

The current flow J out of a point at the box in the direction of the normal vectorn is given by

n · J = σn · E = −σn · ∇u = −σ∂u

∂n. (4.3)

At the insulating hull this current is zero as required by the boundary conditions.At the anode, the total current is given by

Ia = −σ

∫

Aa

∂u

∂ndS, (4.4)

where Aa is the total area of the anode. Similarly at the cathode,

Ic = −σ

∫

Ac

∂u

∂ndS, (4.5)

4.3 Current flow out of the box 69

−5 0 5 10 15−0.1

−0.05

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

y

u

N=124N=496N=1984

4 6 8 10 12 140

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

x

u

N=124N=496N=1984

Figure 4.8: Variation of the potential along x = 4.5 for −4.0 ≤ y ≤

14.0 (above) and along y = 3.0 for 4.10 ≤ x ≤ 14.0 (below) in theplane z = 3.


4 4.2 4.4 4.6 4.8 50.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

x

u

N=124N=496N=1984

Figure 4.9: Behaviour of the quadrature rule near the box,when using Ng = 16 quadrature points. The box surface islocated at x = 4.0 and the potential there is 0.5.

where Ac is the total area of the cathode. In discretised form

Ia = −σ∑

j

∫

∂Ωj

[

∂u

∂n

]

j

dS ; ∂Ωj ∈ Aa, (4.6)

= −σ∑

j

Ajq(j) ; Aj ∈ Aa, (4.7)

where Aj is the area of triangle j at the anode and q(j) = [∂u/∂n]j since we areusing constant elements.

Similarly at the cathode

Ic = −σ∑

j

Ajq(j), Aj ∈ Ac. (4.8)

For our model problem we have computed the above two values and obtained

4.3 Current flow out of the box 71

4 4.2 4.4 4.6 4.8 50.2

0.25

0.3

0.35

0.4

0.45

0.5

x

u

Ng=3Ng=7Ng=16

Figure 4.10: Behaviour of the quadrature rules near the boxfor N = 1984 elements. At the box surface, which is locatedat x = 4.0, u = 0.5.

the results shown in Table 4.5.

N Ia/σ Ic/σ ∆|I|

124 2.0201 -2.2172 0.1971496 2.2399 -2.2882 0.04831984 2.3875 -2.4485 0.0610

(a) ua = −uc = 0.5


124 1.0100 -1.1086 0.0986496 1.1199 -1.1441 0.02411984 1.1937 -1.2242 0.0305

(b) ua = −uc = 0.25


124 0.5050 -0.5543 0.0493496 0.5600 -0.5720 0.01211984 0.5969 -0.6121 0.0153

(c) ua = −uc = 0.125


124 2.1989 -2.1989 0496 2.2627 -2.2627 01984 2.4195 -2.4195 0

(d) Symmetric case, ua = −uc = 0.5

Table 4.5: Current flow out of the box at the electrodes for different valuesof ua and uc in (a), (b) and (c) for the model. In (d) the two electrodes aredirectly opposite each other on opposite sides of the box.

We note from Table 4.5 that the current flowing at the anode is not the same asthat flowing at the cathode. We may be tempted to think that the difference isdue to numerical approximations but this may not be the reason because for thesymmetric case, the two values are the same.


4.4 Examples for other electrode positions

How does the solution behave if we have a different distribution of electrodes?That is what we test in this section. In practice we seek a distribution of elec-trodes that gives an optimum protection.

Figure 4.11 is a plot of the potential obtained when we have two anodes onthe same side at a potential of 0.25 each and a cathode at -0.5 on another side.In Figure 4.12, the two anodes are on opposite sides of the box. In this figure

−0.3

−0.25

−0.2

−0.15

−0.1

−0.05

0

0.05

0.1

0.15

−50

510

−50

510

15−0.4

−0.2

0

0.2

0.4

xy

u

Figure 4.11: The potential (left) and normalised field lines (right)corresponding to two anodes on the same side and a cathode.

we also note the good symmetry properties of the solution. It is good to notefrom these examples that the numerical solution is symmetric for symmetricboundary conditions. The behaviour of the potential and the electric field linesfollow similar trends to the ones discussed above. These results show that our

−0.3

−0.25

−0.2

−0.15

−0.1

−0.05

0

0.05

0.1

0.15

−4 −2 0 2 4 6 8−4

−2

0

2

4

6

8

10

12

14

x

y

Figure 4.12: The potential (left) and normalised field lines (right)corresponding to two anodes on opposite sides and a cathode.

4.5 Richardson extrapolation 73

solution method can be used for any distribution of electrodes. In all cases thepotential obtained falls to zero monotonically away from the electrodes.

4.5 Richardson extrapolation

In this section, we seek to analyse the errors in our simulations. We want toapproximate the order of the error in our approximation and try to improve ourestimates using Richardson extrapolation. If we have the approximate resultsfor two different values of the discretisation parameter h, the mesh spacing, andwe know the order of the error in the approximation as h goes to zero, then wecan extrapolate results from coarse grids to higher order accuracy.

In the above simulations, we choose h and then make an approximation to u.That is, for some constant C and a positive integer p, we have

u = u(h) + Chp + O(hp+1), (4.9)

and for h = h/2,

u = u(h/2) + C(h/2)p + O(hp+1). (4.10)

Multiplying equation (4.10) by 2p and subtracting equation (4.9) gives

(2p − 1)u = 2pu(h/2) − u(u) + O(hp+1),

⇒ u =2pu(h/2) − u(h)

2p − 1+ O(hp+1).

Define

u(h) :=2pu(h/2) − u(h)

2p − 1, (4.11)

then we have

u = u(h) + O(hp+1). (4.12)

The error in (4.12) is of order p + 1 compared to the one of order p in (4.9) andtherefore we expect a better approximation of u. However, in our case we still


do not know the value of p. To calculate p, we consider the following threeapproximations of u:

u = u(h) + Chp + O(hp+1), (4.13a)

u = u(h/2) + C(h/2)p + O(hp+1), (4.13b)

u = u(h/4) + C(h/4)p + O(hp+1). (4.13c)

Subtracting (4.13b) from (4.13a) and rearranging the result we get

−(u(h) − u(h/2)) = Chp(1 − 1/2p) + O(hp+1). (4.14)

Similarly from (4.13b) and (4.13c) we get

−(u(h/2) − u(h/4)) = (Chp/2p)(1 − 1/2p) + O((h/2)p+1). (4.15)

Dividing the two results gives

u(h) − u(h/2)

u(h/2) − u(h/4)=

Chp(1 − 1/2p) + O(hp+1)

(Chp/2p)(1 − 1/2p) + O((h/2)p+1)

=1 − 1/2p + O(h)

(1/2p)(1 − 1/2p) + O(h)

→ 2p (h → 0). (4.16)

Hence p can be estimated from the relation

2p =u(h) − u(h/2)

u(h/2) − u(h/4). (4.17)

For the example in Section 4.1 we have computed the values of p using

p = log

u(h) − u(h/2)

u(h/2) − u(h/4)

/ log 2. (4.18)

For the example problem f = 1/r in Section 4.1, we have calculated the valuesof p in equation (4.18) for h = 2. The results are shown in Table 4.6 for someselected points in the box surface, R is the quotient on the right hand side of(4.17). The results seem to indicate that the value of p is 2. However, we cannotmake any conclusions since there is a significant variation in the values andsome others are even undefined. May be the error does not show the asymptoticbehaviour in (4.9) yet.

4.5 Richardson extrapolation 75

Point R p

(3.333, 6.667, 0) 5.609 2.488(2.667, 7.333, 0) 4.056 2.020(3.333, 8.667, 0) 4.391 2.135(2.667, 9.333, 0) 3.971 1.990(1.333, 6.667, 6.000) 4.002 2.001(0.667, 1.333, 6.000) 4.424 2.145(1.333, 2.667, 6.000) 4.063 2.022(0.667, 3.333, 6.000) 5.642 2.496(1.333, 4.667, 6.000) -2.630 -(0.667, 5.333, 6.000) 2.006 -2.318(1.333, 6.667, 6.000) 9.887 3.306(0, 6.667, 5.333) 3.351 1.745(0, 9.333, 4.667) 3.765 1.912(0, 8.667, 5.333) 4.009 2.003(4.000, 1.333, 0.667) 4.044 2.016(4.000, 0.667, 1.333) 3.803 1.927(4.000, 3.333, 0.667) 3.265 1.707

Table 4.6: Some values of p from equation (4.18).

Chapter 5

Conclusions andRecommendations

In this chapter we conclude what has been done in this thesis. We reflect onboth our work and the original problem and make recommendations for furtherresearch.

5.1 Conclusions

In this thesis we present a detailed account of cathodic protection. We makea distinction between an active (ICCP, Impressed Current Cathodic Protection)and a passive (using sacrificial anodes) cathodic protection system. Matters per-taining with different boundary conditions for the ICCP are also discussed.

We briefly but precisely present the BEM formulation for both the exterior andthe interior potential problems. By use of a problem for which the analyticalsolution is known, we show that the boundary element method is an efficientmethod for calculating solutions to potential problems. The degree of accuracydepends on the type of boundary conditions, that is, Dirichlet, Neumann ormixed boundary conditions; the first one showing the best results.

After testing our method with the example problem, we calculate the potentialand the electric field distribution due to an ICCP system. We demonstrate thevariation of the potential in both three-dimensional and two-dimensional plots.

78 Chapter 5. Conclusions and Recommendations

Our experiments show that for a symmetric problem the potential distributionis also symmetric and so is the current distribution. The potential falls mono-tonically away from the ship. The results also show that, with linear boundaryconditions, the magnitude of the current flow at the anode is not equal to thatat the cathode. However, the two are the same in case of symmetric boundaryconditions.

5.2 Recommendations

Solving the ICCP problem still remains a rich area for scientific research. Al-though we assume linear boundary conditions in our model problem, in prac-tice, for a specified current, the actual potential at the electrode surfaces is givenby polarisation curves which are usually nonlinear. These may further be ad-justed to accommodate the fact that seawater is not a homogeneous continuumeverywhere and at all times. For example, waters near volcanic areas and nearriver mouths are highly saline more than other parts of the sea, and salinity is avery important catalyst for corrosion. Inclusion of these factors into the solutionof the problem by the introduction of polarisation curves will help develop acost-effective cathodic protection system.

The original problem consists of a semi-submerged ship structure. Thus thecomplete solution is that which takes into account the three layers of air, seawa-ter and the seabed. This remains a good case to look at in future research work.In the literature, it is argued that corrosion is more on parts of the surface nearthe seawater-air interface due to the high content of chloride ions and oxygenin this region. The shapes of the ships are usually highly irregular. In futureresearch, it is interesting to try a model whose shape is as close to that of theactual ship as possible.

We have seen that we can include more electrodes in different positions in ourmodel problem and still obtain a well-behaved solution. This fact can be ex-ploited by including more and more portions of the box’s surface as the cathodein the model problem to mimic the vast size of the ship’s surface.

In our discretisations we use uniform elements allover the object’s surface. Thismay not be necessary especially in the case of large models. A finer grid may beused in the critical regions like areas near sharp edges where the integrals mayapproach singularity. More elements may also be used on and around the elec-trodes than in other parts so as to utilise more information from the boundary

5.2 Recommendations 79

conditions on these smaller surfaces.

To save on computation time and memory requirements, symmetric boundaryconditions can be used for the model problem and then the solution can be cal-culated using only half of the model’s surface. The solution on the other halfwill be a mirror image of the one obtained.

Experimenting more on Richardson extrapolation for the BEM also still remainsan interesting piece of work.

Appendix A

Gauss integration for triangulardomains

Gaussian quadrature is often used when we encounter difficult integrations likethe ones required in the formation of the matrices involved in the Boundary El-ement Method. Usually we choose to use triangular elements and in this casethe quadrature integration is carried out on the natural triangle ABC shown inFigure A.1.

A

B

C

ξ

ξ1

ξ2

ξ3

A1

A2

A3

(1, 0, 0)

(0, 1, 0)

(0, 0, 1)

Figure A.1: The natural triangle and coordinates.

To define the natural coordinates ξ1, ξ2 and ξ3 for a point ξ, we subdivide thearea A of triangle ABC into the areas A1, A2 and A3. Then

ξ1 :=A1

A, ξ2 :=

A2

A, ξ3 :=

A3

A. (A.1)

82 Chapter A. Gauss integration for triangular domains

Observe that, since A1 + A2 + A3 = A,

ξ1 + ξ2 + ξ3 = 1. (A.2)

The integration of a function f(ξ1, ξ2, ξ3) over the triangle is performed using aGaussian quadrature such that

∫

A

f(ξ1, ξ2, ξ3) dA = A

Ng∑

i=1

wif(ξi1, ξ

i2, ξ

i3), (A.3)

where wi is the Gaussian weight corresponding to the Gaussian point (ξi1, ξ

i2, ξ

i3)

[Dunavant, 1985]. Note that, actually although the triangle is portrayed in coor-dinates (ξ1, ξ2, ξ3), we only have two degrees of freedom corresponding to ξ1, ξ2

and ξ3 = 1 − ξ1 − ξ2 and the natural triangle ABC is equivalent to the standardunit triangle in two dimension [Pozrikidis, 2002]. Integration in the spatial coor-dinates will involve a transformation between a plane in space and the naturalcoordinate system. This is what we did in (3.83) on page 51. Table A.1 givesthe points ξ1, ξ2 together with the corresponding weights of integration for theorders 3, 6, 7, 16 and 19 from [Bonnet, 1995, page 358]. In the table, the multi-plicity M indicates the number of distinct Gaussian points with the same weightobtained by a permutation of the given point.

83

NG ξk1 ξk

2 wk M

3 0.166666666666667 0.166666666666667 0.166666666666667 3

6 0.445948490915965 0.445948490915965 0.111690794839005 30.091576213509771 0.091576213509771 0.054975871827661 3

7 0.333333333333333 0.333333333333333 0.112500000000000 10.470142064105115 0.470142064105115 0.066197076394253 30.101286507323456 0.101286507323456 0.062969590272414 3

16 0.333333333333333 0.333333333333333 0.072157803838893 10.081414823414554 0.459292588292722 0.047545817133642 30.898905543365937 0.050547228317031 0.016229248811599 30.658861384496479 0.170569307751760 0.051608685267359 30.008394777409957 0.728492392955404 0.013615157087217 6

19 0.333333333333333 0.333333333333333 0.048567898141398 10.020634961602525 0.489682519198737 0.015667350113570 30.125820817014129 0.437089591492935 0.038913770502388 60.910540973211094 0.044729513394452 0.012788837829349 30.036838412054736 0.741198598784498 0.021641769688645 6

Table A.1: Gauss points and weights for integration over the triangleξ1 ≥ 0, ξ2 ≥ 0, 1 − ξ1 − ξ2 ≥ 0.

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