Finite Elementet Lecture Fluid (Good)

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Lecture Notes

Theory of the Finite Element Method

Oliver Heidbach, June 2004

Beta Version 0.97

Contents1. Introduction 3

2. Stating the problem 7

2.1. Constitutive equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72.1.1. Linear elastic rheology . . . . . . . . . . . . . . . . . . . . . . . . 72.1.2. Linear viscous rheology . . . . . . . . . . . . . . . . . . . . . . . . 8

2.2. Equation of motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92.2.1. The Navier-Stokes equations . . . . . . . . . . . . . . . . . . . . . 112.2.2. Equation of equilibrium . . . . . . . . . . . . . . . . . . . . . . . 11

2.3. Mathematical classification of linear partial differential equations of 2ndorder . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

2.4. Physical classification of linear partial differential equations of 2nd order . 13

3. Approximation of the PDE solutions 16

3.1. The Variational Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . 173.2. Proof of the Variational Principle . . . . . . . . . . . . . . . . . . . . . . 173.3. 2-D example for the Variational Principle . . . . . . . . . . . . . . . . . . 193.4. Ritz Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203.5. 2-D example for the Ritz Method . . . . . . . . . . . . . . . . . . . . . . 213.6. 1D-Example for the Ritz Method with numbers . . . . . . . . . . . . . . 233.7. Weighted residuals and Galerkins Method . . . . . . . . . . . . . . . . . 243.8. 2-D example for the Galerkin Method . . . . . . . . . . . . . . . . . . . . 26

4. Numerical solution - The Finite Element Method 28

4.1. First step into the world of Finite Elements . . . . . . . . . . . . . . . . 284.1.1. Transformation for a linear 1-D element . . . . . . . . . . . . . . 314.1.2. Transformation of the derivatives for a linear 1-D element . . . . 33

4.2. 1-D Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

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5. Finite Element Method - Engineering approach 38

5.1. Engineering approach for the 1-D example . . . . . . . . . . . . . . . . . 38

6. Special topics on the Finite Element Method 41

6.1. Do we need shape functions? . . . . . . . . . . . . . . . . . . . . . . . . . 41

6.2. More definitions of Finite Elements . . . . . . . . . . . . . . . . . . . . . 416.2.1. 1D Finite Element with a quadratic polynomial . . . . . . . . . . 416.2.2. Linear transformation for Triangle Elements . . . . . . . . . . . . 426.2.3. 2-D triangular elements with a linear polynomial . . . . . . . . . 43

6.3. Classification of the various element types . . . . . . . . . . . . . . . . . 44

A. Mathematical rules and equations 45

A.1. Positive definite differential operator . . . . . . . . . . . . . . . . . . . . 45A.2. Calculus rules for the first variation . . . . . . . . . . . . . . . . . . . . . 45A.3. Integration by parts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45A.4. Gauss integral equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46A.5. Stokes integral equation . . . . . . . . . . . . . . . . . . . . . . . . . . . 46A.6. Stress definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46A.7. Strain tensor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

B. List of useful books and papers 50

B.1. Books and paper quoted in the text . . . . . . . . . . . . . . . . . . . . . 50B.2. Books on FEM for further reading . . . . . . . . . . . . . . . . . . . . . . 50B.3. Papers on FE-model in Geoscience . . . . . . . . . . . . . . . . . . . . . 51

C. English - German dictionary of mathematical expressions 57

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1. Introduction

These notes are for students from geoscience who are interested in a theoretical back-ground of the Finite Element Methode (FEM) for geodynamic modeling. Due to thewide range of related topics which are involved in this type of modeling (tectonics, rhe-

ology, linear algebra, function analysis etc.) some basic knowledge and understandingin these fields is assumed.

The FEM is not a straight forward method to explain. There is unfortunately the needof lots of mathematics. However, with these notes I do not intend to cover the methodcompletely nor do I try to be precise in the sense of a strict and pure mathematicalapproach. But these shortcomings have to be accepted in order to make the principlesof the method as transparent as possible.

For a more advanced introduction of the FEM lots of books are recommended in thereference list in the appendix. These notes only provide you with the basics of theFEM. They do not give a complete overview and are therefore not an excuse for a cross

reading in professional textbooks, as for instance in the FEM-bible by Zienckiewicand Tayler (1994a and 1994b) where you will find more than 1400 pages in two volumes!This will give you an idea of the complexity and strength of the method. More goodtextbooks with different approaches to the FEM are Altenbach and Sacharov (1982),Kammel et al. (1990) and Schwarz (1991). In these lecture notes the aim is to draw apicture of the main steps creating a geodynamic model with FEM. Easy understandabletwo-dimensional problems using the partial differential equation for the equilibrium offorces (Poisson-equation) with linear elastic rheology will support and accompany thetheory.

In most cases the geoscientist will use a commercial FEM-code, the so-called solver(e.g. MSC MARC, ABAQUS, ANSYS, NASTRAN), which is, unfortunately, a blackbox. Alternatively you can use an open public code, which is to a certain extent writtenby another geoscientist (e.g. Bird and Kong, 1994). These are often poorly documentedand limited to a specific type of problems. A further development is always a very time-consuming task. The alternative is using a commercial FE-software package, which isoften provided with a graphical user interface, containing the a pre- and a post-processor(e.g. MSC MENTAT, PATRAN, ABAQUS CAE, Hypermesh von Altair) and the solver.The pre-processor is needed in order to construct the geometry of the model and itsdiscretization, the post-processor is needed for the visualization of the model results.The core of the package is always the solver, which contains the mathematical algorithmfor the numerical solution procedures (numerical integration, calculation of the inverse

matrix etc.). The reliability of such a software package is high (also the original prices),but the user is always tempted to use the software in a non appropriate way. Solvabilityor mathematical convergence of the numerical problem does not imply that the resultrepresents the observed process, i.e. the process which has to be modeled. The qualityand plausibility strongly depends upon the assumptions and the model input providedand defined by the user. Therefore the results have to be controlled and critically assessedby the user.

Simulation of an observed physical process is a sequence of consecutive steps. The

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first one is always the mapping of the physical process on a model process. The nextones are: finding a solution and a time-spatial description which often implies numericalmethods. An analytical solution is in the most cases not available. The basic idea of anumerical approximation of the (not available) analytical solution is the transformationof the continuous model description into a discrete one. The equation system will then

be solved by a computer with the chosen algorithm (FEM, Finite Differences, BoundaryElement Method ....) Therefore, the simulation, i.e. the modeling procedure can bedivided into three main steps:

Definition of the model process and its geometry

Discretization of the continuum

Application of a numerical algorithm

Therefore the next chapter after this introduction describes the basic differential equationwe have in geoscience and its classification into several groups. Also, some formulas

and statements of rheology will be given. The third chapter covers the mathematicaltreatment of the differential equations before the numerical algorithm of FEM can beapplied. The following chapter four gives the introduction into the FEM itself. In theappendix you will find some useful explanations, some mathematical rules which are usedin the text and an English-German translation of the main mathematical vocabulariesfor your convenience.

But before starting with the mathematics, I would like to encourage a discussionon the term model before we enter into the world of maths. I strongly recommendto start any discussion about a model and its results with a short definition anddescription of the main features of the model type (simplifications, assumptions, type

of boundary conditions, used algorithm and also the visualization procedure) beforediscussing the (geo-)physical meaning, value or result of the model. This will preventlots of misunderstandings. Setting up a common language is always essential and veryhelpful. There is a wide range and definition of the term model or extended terms likekinematic model or physical model in all branches of Geoscience. The following isan incomplete list of different model types:

Tectonic model

Looking at faults (active and inactive ones) and taking into consideration geophys-ical and geological observations and measurements (gravity, heat flow, seismicity,geological maps, fluvial patterns etc.) a tectonic model can be developed. It does

not necessarily give a value as a result (compare with the kinematic model whichcomes somewhat close to this approach). It is more of an integrative approachwhich attempts to bring into accordance with a wide range of observed phenom-ena. (e.g. Sperner et al., 2001)

Kinematic model

A model which only looks for translation and rotation of rigid plates as e.g.NUVEL-1A from deMets et al. (1994). No internal deformation of the lithosphericplate is assumed, i.e. a rigid body (Euklid) is applied.

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Dynamic model

A difficult term and I have, so far, found three different possibilities of an expla-nation from literature: The first uses the term dynamic in models where forces areboundary conditions (e.g. slab pull, ridge push). The second definition is accord-ing to the time-dependence of the model. A pure elastic or elastic-plastic model is

completely time independent and therefore not dynamic, but a model with viscousrheology is time-dependent and can therefore be called dynamic. A third strictdefinition says, that only models which imply acceleration can be called dynamic.In summary: be careful if somebody talks about a dynamic model. I dare notdecide which definition is the correct one.

Mathematical model

Often used in geodesy where a vast amount of observation data of the field variablewith a high accuracy is available. The mathematical model describes the field byreducing the error towards the observed values (e.g. least square method). Thismathematical model does not necessarily imply any physics. It aims only at theperfect mathematical description of the observed field variable data (see figure1).In general the model parameter in a geodetic (mathematical) model is estimatedthrough the observed values (field

variable) by inversion (e.g. least square method, where the model parameter arethe unknown coefficients of the mathematical model). In geophysics and geologythe model parameters are chosen in order to get the field variable as a result of themodel (forward modeling). Of course inverse models are also applied in geologyand in geophysics and vice versa forward model are applied in geodesy.

Figure 1: Sketch for the forward and inverse modeling procedure.

Forward model

This term is used for all modeling approaches where the physical properties andthe boundary conditions of the model are given. The field variable, e.g. the

displacement field, is calculated from the model. In figure (1) this would be themodeling procedure from the right side to the left. The result of the model is thencompared with observed values.

Inverse model

In contrast to the forward model in the inverse model the field variable has beenobserved (e.g. gravity or displacement by GPS) and the model parameters arevaried in order to find the best fit for the observed field variable. Accordingto figure (1) this would be the procedure from the left to the right side. The

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parameters of this mathematical model are varied as long as the average error isminimized.

Numerical model

Any model which uses a numerical (discrete) method due to the lack of availabil-

ity of an analytical (continuous) solution. Methods are Finite Differences, FiniteVolumes, Finite Elements, Boundary Element etc. .

Physical model or analogue model

This is a model which you can actually touch. For instance a sandbox modelconsisting of sirup, plasticine and sand. But this term can also refer to yoursimplification (your model) of the observed phenomena in nature. E.g. back arcspreading behind subduction zones is observed and the physical model which couldgive an explanation is the roll-back of the subduction zone. This physical modeldoes not need to consist of formulas, but is only a picture in your mind. Thishas to be done if you want to convert your physical model into a mathematical

formulation in order to solve it e.g. with an numerical algorithm.

Static model

This term is often used in the commercial FEM software packages. This definitionis based on the type of solving algorithm used in the model. If the problem is pureelastic it is called static model, because the equation of the static equilibrium offorces (accelerations are neglected) has to be solved.

This list is definitely not complete and you will probably find lots of other deviatingdescription in other text books, but it might help you to keep the problem in mindwhenever you find yourself in a discussion about your model with another geoscientist.

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2. Stating the problem

2.1. Constitutive equations

The following constitutive equations are taken from the textbook Rheology of the

Earth by Ranalli (1995). Rheology is Greek1

and can be translated with scienceof flow. Hence rheology describes the deformation (flow) of matter under the presenceof forces. The two most prominent linear rheological laws, also called constitutive equa-tions, are Hookes law for elastic deformation and Newtons law for viscous deformation.They relate the stress tensor ij to the deformation tensor ij. A combination of thesetwo can describe many of the deformation processes observed in the lithosphere.

The stress tensor ij describes the stress state of any point in a continuous body.The strain tensor kl describes the deformation, i.e. the change of displacement withspace within a continuous body. The relation between the applied forces (or the appliedstresses, since stress is measured in force per area with the measure unit Pascal) isdescribed by rheological laws.

2.1.1. Linear elastic rheology

The theoretical model of perfect elasticity postulates that the components of strain atany point within an elastic medium are homogeneous linear functions of the componentsof stress. The most general description is therefore

ij = Cijklkl . (1)

where Cijkl represent the 81 elastic parameter. Assuming isotropy, i.e. the physicalproperties do not depend on the direction, this number can be reduced to the two Lame

constants and . With the Kronecker symbol and the deviatoric part

ij of the straintensor, Hooke law for an isotropic material can be stated as

ij = kkij + 2

ij . (2)

Hookes law is time-independent, i.e. the reaction of the pure elastic material isinstantaneous. As long as the applied forces do not change, the deformation will notchange either. The stress tensor can be divided into two parts: The diagonal componentsof the stress tensor (11, 22 and 33) which describes the lithostatic pressure (see figure2) and the shear components (off-diagonal components). Since only this deviatoric part(denoted by ) of the stress tensor produces deformation2 it is convenient to extractthis part and we obtain

ij = 2

ij (3)

1 - everything flows. This statement is from Herakleitos of Ephesos around 500 v. Chr.,who was one of the first natural philosophers trying to explain the processes in nature)

2This statement is somewhat tricky and implies that a pure volume change of a material is notcalled deformation. When 11 = 22 = 33 is applied to rock material, the volume will decreaseassuming the rock is compressible. But this deformation is not of interest in geodynamic processes.Therefore the term deformation is mostly used in the context of deformation due to deviatoricstresses which change the shape of the rock

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Instead of the Lame constants the Young modulus E and the Poisson number is oftenused:

E =(3 + 2)

+

= 2( + )

=E

2(1 + )(4)

where is called the shear modulus often given as G. The range of E and G for Granitefor example is in the order of 10-100 GPa (depends on temperature, pressure, textureetc.) whereas is around 0.3 (maximum is 0.5 for an incompressible material as, forinstance, water).

2.1.2. Linear viscous rheology

A perfect linear viscous material (linar Newton fluid) can be stated as

ij = pij + ij + 2ij . (5)

where p is the pressure, the change in volume, a material constant and is theviscosity. Since a viscous deformation process is time-dependent the Newton law statesthe relationship between the applied forces and the deformation rate. When the materialis assumed to be incompressible ( = 0) the linear Newton law becomes

ij = pij + 2ij

For the shear component ij of the stress tensor ij we therefore obtain

ij = 2ij . (6)

The dot over the strain and shear strain tensor is the time derivative which gives usthe strain rate tensor and the shear strain rate tensor respectively. The strain tensor isdefined as

ij =1

2ui

xj+

uj

xi(7)

and for the strain rate tensor we receive

ij =

t

1

2

ui

xj+

uj

xi

=1

2

vi

xj+

vj

xi

(8)

where v is the velocity.

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2.2. Equation of motion

The equations of equilibrium are applied for a description of an continuous body to be inequilibrium, i.e. the equilibrium of the forces which are acting on and in the body. Foran equilibrium the resultant moment about any axis must vanish. This condition implies

that the sum of forces in any coordinate direction must vanish. For the description ofthe forces acting on the surface of a body the Cauchy stress tensor ij is used. Accordingto figure (2) this stress tensor is defined as

ij =

11 12 1321 22 23

31 32 33

(9)

where 11, 22, 33 are the normal stresses and 12, 13, 23, 21, 31, 32 are the deviatoricstresses3. With reference to figure (2) the acting force on any surface of the parallelepiped

x3

x2

x1

dx3

dx2

dx1

32

33

31

22

23

21

12

13

11

Figure 2: The figure shows the three stress components of the stress tensor ij on thethree faces of an infinitesimal parallelepiped. The volume of the infinitesimal

parallelepiped is defined by its side lengthdx

1, dx

2 anddx

3 in a cartesiancoordinate system

is given by the relevant stress component multiplied by the area of the face. Since thereare two parallel faces in each direction of the coordinate system and assuming thatthe stress components meet certain continuity conditions (first and second derivative

3Following figure (2) the first subscript i of the stress tensor denotes the coordinate axis to which thesurface is normal and the second subscript j the direction to which the stress is applied to.

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towards space must be continuous) stresses (force per area) acting in x1-direction are

11dx2dx3 +

11 +

11

x1dx1

dx2dx3

21

dx1

dx3

+ 21 + 21x2 dx2dx1dx3 (10) 31dx1dx2 +

31 +

31

x3dx3

dx1dx2 .

Rearranging this equation gives11

x1+

21

x2+

31

x3

dx1dx2dx3 (11)

These surface forces have to be in equilibrium with the body forces. With the densityof the material and X1 the body force per unit mass the body forces in x1-direction is

X1dx1dx2dx3. Therefore the equation of equilibrium in x1-direction is11

x1+

21

x2+

31

x3+ X1 = 0 (12)

The same procedure for the x2 and x3-direction gives

12

x1+

22

x2+

32

x3+ X2 = 0 (13)

13

x1+

23

x2+

33

x3+ X3 = 0 (14)

which can be summarized with index notation as

ij

xi+ Xj = 0 (15)

where i, j = 1, 2, 3. The stress tensor must be symmetric, i.e. ij = ji since the shearstresses have to cancel each other in equation (10). Otherwise a net torque would acton the body and the equilibrium condition would not be fulfilled. The equations ofequilibrium state a steady state field problem, since the field variable (for instance thedisplacement field) is time-independent.

If acceleration aj can not be neglected,

ij

xi+ Xj = aj (16)

expression (16) is called the equation of motion. Since the field variable (e.g. displace-ment u) is dependent on time (the acceleration is the second derivative towards time)the equation of motion is a non-steady state field problem, whereas the equations ofequilibrium are time-independent and therefore a steady-state field (stationary) prob-lem.

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2.2.1. The Navier-Stokes equations

When the increments of the velocity vj are small the acceleration can also be written inthe following form:

aj = vj = uj =Dvj

Dt =D2uj

Dt2 =2uj

t2 (17)

where D is the total derivative, vj the velocity and t the time. By inserting the definitionof the compressible Newton fluid from expression (5) into expression (16) and assumingsmall velocity increments we receive

xi(pij +

ij + 2ij) + Xj = 2uj

t2. (18)

These equations of motion are called the Navier-Stokes equations (for a compressiblefluid with linear Newton viscosity). Neglecting the acceleration, which is reasonable for

most of the tectonic processes, and assuming incompressibility ( = 0) we receive

xi(pij + 2ij) + Xj = 0 . (19)

Multiplying expression (19) with 1 and inserting expression (8) for ij leads to

1

p

xi+

1

xi

vi

xj+

vj

xi

+Xj = 0

1

p

xi+

1

2 uj + Xj = 0 (20)

2.2.2. Equation of equilibrium

A body in equilibrium implies that accelerations can be neglected in expression ( 16).By inserting the Hook law for the shear component from expression (3) and expression(7) for the strain tensor we receive

xi(2ij) + Xj = 0

uj

2

x2

i

+ Xj = 0

2 uj + Xj = 0

uj + Xj = 0 (21)

assuming that , the shear modulus (also denoted by G in the literature) is homogeneous.This is a Poisson equation which is a linear partial differential equation of 2nd order.

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2.3. Mathematical classification of linear partial differential

equations of 2nd order

Most of the differential equations in geoscience are linear partial differential equations(PDEs) of 2nd order. In two dimensions they can be stated in a general form as

A2u

x21+ 2B

2u

x1x2+ C

2u

x22= F (22)

where A , B , C , F and u are functions of the coordinates x1, x2. The mathematicalclassification differentiates between parabolic, hyperbolic and elliptical PDEs4 accordingto the value of D = AC B2.

parabolic D = 0

hyperbolic D < 0 (23)

elliptic D > 0

According to their physical meaning the PDEs are classified into two groups: the time-independent steady state field problems and time-dependent non-steady state field prob-lems.

Physical description of the observed

geophysical phenomena through

continuums mechanics

Partial Differential Equation (PDE)

linear, non-linear, order

Linear PDE of 2nd order

Classification into:

Parabolic PDE

(diffusion equation)

Hyperbolic PDE

(wave equation)

Elliptical PDE

(Poisson equation)

plus boundary conditions

Correct stated problem:

Existance, unambiguousness

Applicability of

numerical methods

Figure 3: Flowchart of the classification of linear Partial Differential Equations (PDE)of 2nd order and their general solving scheme.

4This comes from the planar analytical geometry where ax2 + 2bxy + cy2 + dx + ey + f = 0 describesa parabola, hyperbola or ellipse in a case if d = ac b2 = 0, < 0, or > 0

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2.4. Physical classification of linear partial differential equations of

2nd order

The most general description of linear partial differential equations of 2nd order in atwo-dimensional problem with anisotropic material behavior is

x1

a1(xi, t)

u

x1

+

x2

a2(xi, t)

u

x2

= f(xi, t) + b(xi, t)

u

t+ c(xi, t)

2u

t2(24)

where u is either a scalar (e.g. temperature) or a vector (e.g. displacement) and thefunctions a1(x1, x2) and a2(x1, x2) describe the material properties (i, j = 1, 2 accordingto the spatial dimensions).

According to figure (4) the PDE is defined in the area S with the border C. By

A

C

n

x1

x2

c

S

Figure 4: Sketch of the surface S where the PDE is described. On the border C, whichis divided into two parts C1 and C2, the boundary conditions are prescribed.n is the outer normal of the border and c is the circular length.

dividing the border C into two segments C1 and C2, various boundary conditions canbe described at the border. The function u(x1, x2) has to fulfill certain values which areprescribed by the functions (c), (c) and (c) of the circular length c.

u(c) = (c) on C1 Dirichlet boundary condition (25)u(c)

n(c)+ (c) u(c) = (c) on C2 Cauchy boundary condition (26)

where n(c) is the outer normal of the border C. The Cauchy boundary condition canbe reduced to the following Neumann boundary condition

u(c)

n(c)= 0 on C2 Neumann boundary condition (27)

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in case of(c) = (c) = 0. Since n(c) is the outer normal of the border C, the Neumannboundary condition denotes that u(c) is not allowed to change perpendicular to theborder C of the surface S.

Also, the boundary conditions are now time-dependent. Additionally an initial value,which describes the state of the field variable at time t0, has to be given. Three varieties

of expression (24) for a general partial differential equation (PDE) can be distinguishedand are well known in geoscience.

Type Mathematical Physical classification

classification Example

c(xi, t) = 0 parabolic PDE non-steady statediffusion equation

b(xi, t) = f(xi, t) = 0 hyperbolic PDE non-steady stateHelmholtz or wave equationin elasto-dynamics

b(xi, t) = c(xi, t)=0 elliptical PDE steady stateequation of equlibriumpotential equation

For steady state field problems the PDE only depends on u and their spatial deriva-

tives, where u is either a scalar (e.g. temperature) or a vector (e.g. displacement). Thefunctions a1(x1, x2) and a2(x1, x2) describe the material properties. In an anisotropicnon-homogeneous material the general expression for a steady-state field problem in twodimensions is:

x1

a1(x1, x2)

u

x1

+

x2

a2(x1, x2)

u

x2

= f(x1, x2) (28)

An geoscientific example is the calculation of the geothermal gradient with depth. Thefield variable is the u = T(x1, x2), the material property is the thermal diffusivity andthe function f(x1, x2 is the heat production due to radioactive decay. Assuming that

is homogeneous expression (28)

a1 = a2 = = 106 (29)

the PDE is the Poisson equation

1062T

x21+ 106

2T

x22= 106 T = f(xi) (30)

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In the case of no heat production (f(x1, x2) = 0) and no dependence in x1-direction(x2 beeing the vertical direction of the lithosphere) the Poisson equation is called theLaplace equation

2T

x22= 0 (31)

With boundary conditions at the bottom of the lithosphere (Tm = 1500 K; meltingtemperature of lithospheric rocks) and the Earth surface (T0 = 280K) the Laplaceequation can be solved by integration.

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3. Approximation of the PDE solutions

In most cases it is not possible to find an analytical solution for the PDE. As a conse-quence only an approximation of the solution can be found by applying an appropriatenumerical algorithm. Each of the examples given in figure ?? needs a discretisation of

the model area. There is always the fight between accuracy and flexibility (capabilityto describe irregular geometries and complicated boundary conditions) of the numeri-cal method. E.g. the accuracy of the Finite Difference method is higher compared tothe FEM, but the flexibility of the FEM is much better. As a preparation before we

Description with a "physical" model

Mathematical Formulation with a

Partial Differential Equation (PDE)linear, non-linear, 2nd order

Application of different

solving schemes

Classification into:

Methods of

weighted residuals

(Galerkin, Collocation)

Boundary Element

MethodFinite Differences

Method

Variational Formulation

Extremal principle

Approximation with

Ritz Method

Finite Element

Method

Finite Element

Method

Observed geophysical phenomena

Figure 5: Flowchart of the classification of Partial Differential Equations (PDE) of 2nd

order and their general solving scheme

apply the Finite Element Method we have to rewrite the PDE. This can be achieved

in two different ways. The variational formulation with subsequent Ritz method or theGalerkin method (method of weighed residuals). Both approaches (Ritz and Galerkin)seek numerical approximations of the wanted solution for the PDE and they both leadto the same set of equations, but still a solution is only found for simple geometries ofthe model area. However both approaches will bring us directly to the world of FiniteElements.

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3.1. The Variational Principle

In contrast to an extremum calculation for functions, i.e. looking for a value of thefunction where the function has an extremum (minimum or maximum), in variationalcalculus one seeks a function which minimizes a functional. A functional is an equation

which is not dependent on coordinates anymore but on the functions. Functions are nowthe variables.Let us consider the functions a1, a2, a3, u, f and g in a given space V R

3. S isthe surface of V, S1 and S2 are parts of S and S1 S2 = S. The functions (s) and(s) describe the boundary conditions and are therefore defined on the surface S of V.The vector n is the outer normal onto the surface S2. The function u(xi) with i = 1, 2, 3which then minimizes the functional

I(u) =

V

1

2

a1

u

x1

2+a2

u

x2

2+a3

u

x3

2

1

2f u2 gu

dV

+ S12 (s) u2(s) (s) u(s) dS (32)also solves the differential equation

x1a1

u

x1

+

x2a2

u

x2

+

x3a3

u

x3

+f u + g = 0 (33)

with the Dirichlet boundary condition u(s) = (s) on S1 and the general Cauchy bound-ary condition on S2

a1u

x1

n1 + a2u

x2

n2 + a3u

x3

n3 + (s) = (s) . (34)

Every differential equation, where the linear differential operator L is positive definite,can always be stated as a variational formulation5. For the special case where (s) =(s) = 0, the general Cauchy boundary condition is called the Neumann boundarycondition.

3.2. Proof of the Variational Principle

In order to solve the variational problem the first variation I(u) of the integral expres-sion I(u) (which is the first term of the Taylor expansions for functionals) has to vanish.

Or in other words, to find a minimum of the functional I(u) This leads to

I(u) =

V

1

2

a1

u

x1

2+a2

u

x2

2+a3

u

x3

2

1

2f (u2) gu

dV

+

S

1

2(u2) u

dS = 0 . (35)

5If L is self-adjoint and elliptic, L is positive definite (see also appendix A). The differential operatorLaplace and the differential operator Nabla operator are positive definite

17


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Due to the condition that these integrals have to vanish for all variations of u of u,the expression under both integrals must vanish, i.e. they fulfill the stated differentialequation (33). If one of the integrals would not equal zero it would be possible to findsuch u that the left hand side of expression (40) would not equal zero. The functionu(xi) which minimizes expression (32) and solves the partial differential equation (33)

with the boundary conditions stated in (34).

3.3. 2-D example for the Variational Principle

Let us assume that = = 0, g = 1, f = 0, a1 = 1, a2 = 2 in a two-dimensionalproblem with the cartesian coordinates x1 and x2. Filling this into expression (33) wewill then receive:

2u

x21+ 2

2u

x22+ 1 = 0 (41)

with the Dirichlet boundary condition u = on C1

and the general Cauchy boundarycondition on C2

u

x1n1 + 2

u

x2n2 = 0 (42)

According to expression (32) the variational formulation of this PDE leads to the fol-lowing integral expression

I(u) =

S

1

2

u

x1

2+2

u

x2

2+u

dS (43)

In order to solve the variational formulation the first variation I(u) of the integralexpression I(u) (which is the first term of the Taylor expansions for functionals) has to

vanish. This leads to

I(u) =I(u)

u=

S

1

2

u

x1

2+2

u

x2

2u

dS = 0 (44)

Applying the calculus rules for the first variation (see appendix A, expression (170)) onequation (44) we receive

I(u) =

S

u

x1

x1u + 2

u

x2

x2u u

dS = 0 (45)

Applying integration by parts (see appendix A, expression(171)) on the first two terms

of the surface integral (45) we receive for every termS

u

x1

x1(u) + 2

u

x2

x2(u)

dA

=

S

x1

u

x1u

dS

S

x1

u

x1

u

dS

+

S

x2

2

u

x2u

dS

S

x2

2

u

x2

u

dS (46)

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Applying the integral sentence of Gauss (see appendix A, expression (176) the twopositive surface integrals of (46) can be changed into a line integral. Equation (46) thenbecomes

S

x1

u

x1

u dS S

x1

u

x1u dS

+

S

x2

2

u

x2u

dS

S

x2

2

u

x2

u dS

=

C

u

x1n1u dC

S

x1

u

x1

u dS

+

C

2u

x2n2u dC

S

x2

2

u

x2

u dS (47)

and by rearranging this leads directly to

= C ux1n1u dC+ C2 ux2 n2u dC

S

x1

u

x1

u dS

S

x2

2

u

x2

u dS (48)

=

C

u

x1n1 + 2

u

x2n2

u dC

S

x1

u

x1

+

x2

2

u

x2

u dS (49)

If expression (49) is substituted in (45) we receive

S

x1

ux1

+

x22 u

x2

+1

u dS

+

C

u

x1n1 + 2

u

x2n2

u dC = 0 (50)

Due to the condition that these integrals have to vanish for all variations of u of u,the expression under the integrals must vanish, i.e. they fulfill the stated differentialequation (41). That is, the function u(xi) which minimizes equation (43) also solves thepartial differential equation (41). But also for the variational formulation of the PDE,where the order of the differential equation is reduced by one, it is difficult, or even

impossible, to find an analytical solution.

3.4. Ritz Method

As stated in the previous chapter an analytical solution is difficult to receive for thePDE and its variational formulation. Therefore an approximation procedure must beapplied. Let us consider a set of linear independent functions

0(xi), 1(xi) . . . , m(xi) . (51)

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As the problem is not solvable in u(xi), i = 1, 2, 3 an approximation uh(xi) as a linearcombination of k(xi) is chosen.

uh(xi) =m

k=0k k(xi)

= 00(xi) +mk=1

k k(xi) . (52)

It is reasonable that 0(xi) fulfills the boundary conditions and that all other j(xi) forj = 1,....,m vanish at the boundary. Substituting this approximation into equation (32)one gets

I(0, . . . , m) =

V

a1

2

mk=0

kk(xi)

x1

2+

a2

2

mk=0

kk(xi)

x2

2+

a3

2

mk=0

kk(xi)

x3

2

f2

mk=0

k k(xi)2

g m

k=0

k k(xi)

dV

+

S

2

mk=0

k k(xi)

2

mk=0

k k(xi)

dS (53)

where 0 = 1 and 0(xi) fulfils the boundary conditions on S. Therefore only the re-maining m functions m(xi) have to be taken into consideration. In order to makethis expression extremal one has to set the first partial differentiations to zero (with

j = 1,....,m because 0 is per definition already known)

I(j) =I

j=

V

a1

mk=0

kk(xi)

x1

j(xi)

x1

+ a2

mk=0

kk(xi)

x2

j(xi)

x2+ a3

mk=0

kk(xi)

x3

j(xi)

x3

dV

V

f

mk=0

kk(xi)j(xi)

+g j(xi)

dV

+ Sm

k=0 kk(xi)j(xi) j(xi) dS = 0 (54)3.5. 2-D example for the Ritz Method

Again as in the 2-D example for the Variational Method ( = = 0, g = 1, f = 0,a1 = 1, a2 = 2) the given PDE is

2u

x21+ 2

2u

x22+ 1 = 0 (55)

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with the Dirichlet boundary condition u = on C1 and the general Cauchy boundarycondition on C2

u

x1n1 + 2

u

x2n2 = 0 . (56)

From expression (54) we receive with the above stated definitions for , , g , f , a1 anda2 for this two dimensional problem

I

j=

S

mk=0

kk(xi)

x1

j(xi)

x1+ 2

mk=0

kk(xi)

x2

j(xi)

x2 j(xi)

dS = 0 (57)

which can be rewritten intoS

mk=1

kk(xi)

x1

j(xi)

x1+ 2

mk=1

kk(xi)

x2

j(xi)

x2

dS

+ S0 0(xi)x1j(xi)

x1+ 2 0

0(xi)

x2

j(xi)

x2 j(xi)dS = 0 (58)

Rearranging expression (58) and taking into account that 0 = 1 is defined in order tofulfill the boundary conditions [0(xi) describes the boundary conditions at the boundarywhere all other function j(xi) vanish] we receive

S

mk=1

kk(xi)

x1

j(xi)

x1+ 2

mk=1

kk(xi)

x2

j(xi)

x2

dS

+

S

0(xi)

x1

j(xi)

x1+ 2

0(xi)

x2

j(xi)

x2 j(xi)

dS = 0 . (59)

This leads to a set of linear equations of the j unknown parameter j

ajk :=

S

k(xi)

x1

j(xi)

x1+ 2

k(xi)

x2

j(xi)

x2

dS (60)

bj :=

S

0(xi)

x1

j(xi)

x1+ 2

0(xi)

x2

j(xi)

x2 j(xi)

dS (61)

or in matrix form

K = (ajk)

c = (1, . . . , m)T

b = (b1, . . . , bm)T

for j, k = 1, , m

K c = b . (62)

From definition (60) it can be seen that the matrix K is symmetric, i.e. ajk = akj .

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3.6. 1D-Example for the Ritz Method with numbers

Let us consider the partial differential equation

ij

xi+ Xj = 0 . (63)

In the 1D case expression (63) reduces according to figure (6) with no body forces to

Ed2u

dx2= 0 (64)

with the given Hooke law in one dimension

= E and =du

dx(65)

dx = E du

where the E-Modul is assumed to be constant in time and space.

x

=10 MPa

0 1

E=50 GPa

0

Figure 6: The bar is fixed to the wall on the left-hand-side. It has a constant YoungsModul E = 50 GPa and is pulled by a force of 0 = 10 MPa on its rightside. The length of the bar is one meter along the x axis.

According to figure (6) this is a bar of length one along the x-axis which is fixed ona wall on its left-hand-side. At the right side the bar is pulled by a force of 0 = 10MPa. The Young Modulus of the bar is constant and has a value of E = 50 GPa. Theboundary condition of the differential equation is u(x = 0) = 0 for the displacement atthe wall. By inserting the given force boundary condition 0 at the position u(x = 1)

into the Hooks law we receive a second boundary condition in terms of the displacement

u(x = 1) =0

E(66)

According to expression (34) the boundary condition at x = 1 of the given 1D problemis

a1u

x1n1 = E

u

x= 0 (67)

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The variational formulation of the differential equation (64) is

I(u) =

10

1

2E

du

dx

2dx 0u(x = 1) (68)

Let us now assume a linear approximation function uh(x) for u(x)

uh(x) = 0 + 1x (69)

uh(x) = 1x

since uh(x = 0) = 0 = 0 due to the given boundary conditions. Inserting this approxi-mation into the variational formulation gives

I(1) =

10

1

2E

d

dx(1x)

2dx 01 (70)

= 1

0

1

2E21dx 01

The first variation gives

I(1) =dI(1)

d1= 0 (71)

=

10

E1 dx 0

= E1 0

1 = 0E

With the given values for E and the boundary conditions we then receive

1 = 0, 0002 (72)

uh(x) = 0, 0002 x

Since the stated linear 1D problem is solved exactly by a linear function, the chosenapproximation uh(x) is the exact solution of the differential equation in expression (64).

3.7. Weighted residuals and Galerkins Method

In case of a positive definite differential operator6 a variational formulation always exists.But for other PDEs this is often not possible. Another approach has to be chosen, inorder to get an appropriate for of an approximated formulation of the solution whichcan be used in a numerical method. This approach is the method of weighted residuals

6this is always true when the PDE is elliptical

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of which the Galerkin7 method is a special case. Again we choose an approximationuh(xi) of the solution u(xi)

uh(xi) =m

k=0k k(xi)

= 00(xi) +mk=1

k k(xi) (73)

where, again, 0(xi) fulfils the boundary conditions. Substituting this approximationinto expression (33) we obtain

R(xi) =

x1a1

mk=0

kk(xi)

x1

+

x2a2

mk=0

kk(xi)

x2

+

x3a3

mk=0

kk(xi)

x3

+ fm

k=0 k k(xi) + g (74)where R(xi) is the residual (defect) of the approximation. If uh(xi) were the exactsolution the residual R(xi) would equal zero. Hence, the nearer R(xi) tends towardzero, the better we can expect the approximation to be. The idea of the weightedresidual method is that the residual is zero, but only in a weighted sense. The weightis provided by the j(xi). Every function j(xi) gives one equation. In this special casewhen the weighting function is the function used in the approximation the method iscalled Galerkin method8.

The set of equations for the m unknown coefficients 1, . . . , m follows from the re-quirement that the sum (=integral) of the product of R(xi) and j(xi) is identical to

zero V

R(xi) j(xi) dV = 0 f or j = 1, . . . , m (75)

Mathematically spoken: If the inner product of two functions u and v is zero

< u, v > =

V

u v dV = 0 (76)

the functions u and v are orthogonal to each other. For our case this implies that theresidual R(xi) should be orthogonal to all test functions j(xi).

< R(xi),

j(x

i) > = V R(xi) j(xi) dV = 0 (77)

That this requirement of orthogonality is good, is illustrated in figure 7. If the residualR(xi) is orthogonal to the function j(xi) it is the smallest.

7Boris G. Galerkin (1871-1945). Galerkin was a Russian engineer who published his first technicalpaper on the buckling of bars while imprisioned in 1906 by the Tzar in pre-revolutionary Russia,In many Russian texts the Galerkin method is known as the Bubnov-Galerkin method. Galerkinpublished a paper using this idea in 1915. The method was also attributed to I.G. Bubnov in 1913.

8If, for instance, the Dirac delta function is used as the weight function the method is called collocation

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R

1

u

uhu1

R

1

u

uh

2u2

u= u11 + u22

R.1+R.2=0

u= u11

R.1=0

Figure 7: The sketch shows how the Galerkin method maintains orthogonality betweenthe the residual vector R and the set of basis vectors j . Two cases for oneand two basis vectors are shown. In case of orthogonality the residual R is thesmallest.

3.8. 2-D example for the Galerkin Method

Again as in the 2-D example for the Ritz Method, ( = = 0, g = 1, f = 0, a1 = 1,a2 = 2) the given PDE is

2

ux21

+ 2 2

ux22

+ 1 = 0 (78)

with the Dirichlet boundary condition u = on C1 and the general Cauchy boundarycondition on C2

u

x1n1 + 2

u

x2n2 = 0 (79)

From expression (74) we receive with the above stated definitions for , , g , f , a1 anda2 for this two dimensional problem

R(xi) = 1 +m

k=0k

2k(xi)

x21+ 2

m

k=0k

2k(xi)

x22(80)

The equations for the coefficients 1, . . . , m are received from the requirement that theintegrals

S

R(xi) k(xi) dS (81)

are identical to zeroS

1 +

mk=0

k

2k(xi)

x21+ 2

2k(xi)

x22

j(xi) dS = 0 (82)

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Application of the partial integration, then applying the integral sentence of Gauss andexchanging of integration and summation as well as taking into consideration that 0 = 1we receive

m

k=1

k Sk(xi)x1j(xi)

x1+ 2

k(xi)

x2

j(xi)

x2 dS+

mk=1

k

S

0(xi)

x1

j(xi)

x1+ 2

0(xi)

x2

j(xi)

x2 j(xi)

dS = 0 (83)

This set of linear equations ofj unknowns parameter j is identical to the set we receivedfor that PDE with the Ritz Method in expression (59). Analog to the Ritz Method wecan then define

ajk :=

S

k(xi)

x1

j(xi)

x1+ 2

k(xi)

x2

j(xi)

x2

dS (84)

bj :=

S

0(xi)

x1

j(xi)

x1+ 2

0(xi)

x2

j(xi)

x2 j(xi)

dS (85)

or in matrix form

K = (ajk)

c = (1, . . . , m)T

b = (b1, . . . , bm)T

for j, k = 1, , m

K c = b . (86)

From definition (84) it can be seen that the matrix K is symmetric, i.e. ajk = akj .

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4. Numerical solution - The Finite Element Method

Since we can not determine an approximation function by applying the Ritz or theGalerkin we now have to apply the Finite Element Method (FEM). This procedure in-cludes the steps of discretization, finding an appropriate approximation function for this

discretization, numerical integration and solving a linear set of equations (i.e. invertinga coefficient matrix).

4.1. First step into the world of Finite Elements

A global approximation function (polynomial, Fourier series etc.), as used in the RitzMethod, has three main disadvantages:

The matrix K = (ajk) is full, i.e. it is difficult to invert.

The approximation function is not very flexible, i.e. a local refinement in critical

areas is almost impossible

An approximation function is difficult to estimate where the geometry of the modelarea A and its boundary conditions are complicated. Especially a non-regulargeometry of the model boundary is difficult to describe.

In order to overcome this mission impossible we subdivide the surface S with its bor-der S into subsurface (see figure 8). These subareas should not overlap each other andshould have easy and smooth borders. In a two-dimensional case triangles and quadran-gles are the most common definitions of subareas (finite elements). Their borders canbe described by linear polygons. Instead of defining a complicated global approximation

model area S

borderof S

Finite Element

nodal point

Discretisation

Figure 8: The model area S with its boundary S is subdivided into subareas. These

subareas can be triangles or quadrangles. These subareas are then called theFinite Elements. The interconnecting points between these Finite Elements atthe corners are called nodal points or in short nodes.

function for the whole model surface S, we now only have to find an approximationfunction which describes the model behaviour in the subarea. We also reduce the conti-nuity and differentiability requirements of this approximation. That means we only needto find a piecemeal continuous and piecemeal differentiable approximation function for

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each subarea Dj with its border Dj. Outside their subareas these local approximationfunctions are not defined. Mathematically spoken we have to fulfill the following twoconditions:

(S S) =

m

j=1

(Dj Dj) (87)

j(xi) =

approximation function , if xi Dj Dj0 , else

(88)

This is the basic idea which leads from the classical Ritz or Galerkin method to theFinite Element Method (FEM). The most convenient local approximation functions forthese subareas are polynomials (linear, quadratic or cubic).

By using a linear approximation of the function u(xi) in a subarea we have to besure that within the chosen subarea the variation of u(xi) can be approximated witha linear polynomial. If this is not the case we either have to choose a finer subdivision(later called mesh refinement) or a higher order of the polynomial approximation (cubicor quadratic) for the subarea. Of course the higher the number of subareas is and/orthe higher the polynomial order is, the higher the computational costs are. But this willbe discussed later. In a two-dimensional example the approximating function uh(xi) forthe subsurface S is

uh(xi) = 0 + 1x1 + 2x2 (89)

uh(xi) = 0 + 1x1 + 2x2 + 3x21 + 4x

22 + 5x1x2 + 6x1x

22 + 7x2x

21 (90)

where the first function is a linear polynome and the second a quadratic polynome.The order of the approximation function is dependent upon the given problem and itsdiscretization.

An important requirement is that continuity of the global solution uh(xi) from oneelement to the neighboring elements must be fulfilled9. Elements which fulfill thesecontinuity requirements are called conform. Besides this conformity requirement it isalso required that the approximation functions do not change their geometrical formsunder transformation (a straight line should remain a straight line after transformation)from one coordinate system to another. In this case the approximation functions are

called (geometric) isotropic.The following step is the basic idea of the Finite Element method. In order to fulfill

the continuity requirements from one element to the next the usage of coefficients i isnot suitable. It would be much more convenient to express the approximation functionwithin the elements (I will use this term from now on in the text instead of the term

9Most of the continuity requirements are also obvious from a physical point of view. If, for instance,uh(xi) describes the displacement field, a non-continuous behaviour from one element to the nextwould produce gaps.

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subarea) through the function values ofuh(xi) at certain points of the elements10. These

points are then called nodal points or nodes of the elements. In order to fulfill continuityfrom one element to the neighbouring one it is sensible to place these nodes at the cornersof an element, i.e. a node is always shared by one or more elements. The values

u(e)k (xi) (91)

where e = 1,...,n is the number of the element,

p = the number of nodes per element and

l, k = 1,...,p the node number of the element

of the approximation function at these nodes are called the nodal values of the nodevariable. With these nodal values the approximation function can be expressed as alinear combination of functions which use the nodal values of the nodal variables astheir coefficients. For a two-dimensional element with p nodes we would then write:

u(e)h (xi) =

pk=1

u(e)k N

(e)k (xi) (92)

Since this expression must be valid for all nodal values u(e)k (xi) the functions N

(e)k (xi)

must have the value one at the node k with the coordinates x(e)i,k and zero at all othersnodes of element e

N

(e)

k (x(e)

i,l ) = 1 for k = l0 for k = l (93)If the nodal values are now globally (all elements) numbered from 1,...,m, the globalapproximation can be written as:

uh(xi) =mk=1

ukNk(xi) (94)

where Nk(xi) is the combination of all element shape functions N(e)k (xi) which have the

value one at the nodal point k with the nodal value uk.

The expression (94) which approximates the wanted function u(xi) has the same shapeas the approximation function we used in the Ritz and Galerkin method. However, thevalues of the unknown coefficients j are now the values of the wanted function u(xi) atthe node k, i.e. uk(xi).

10Sometimes the partial derivatives of the approximation function uh(xi) are taken. E.g. solving thedisplacement field, the partial derivative gives the strain field.

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4.1.1. Transformation for a linear 1-D element

It is sensible to map the approximation function uh(xi) onto a standard interval [0, 1] inthe coordinate space. This mapping is done by defining the transformation formula.For a one dimensional linear approximation function the transformation formula from

the x coordinate system into the coordinate system is

x = xj + (xj+1 xj) (95)

For the linear approximation

(x) = 0 + 1x (96)

we receive in the coordinates

() = 0 + 1[xj + (xj+1 xj) ]

= 0 + 1xj :=0

+ 1(xj+1 xj) :=1

() = 0 + 1 (97)

where xj and xj+1 are the nodal values. Another possibility to represent the straightline is by defining so called shape functions. Since there is two nodes (or two unknowncoefficients) for the linear approximation for the one dimensional problem, we need twoshape functions N1 and N2

N1() = 1 N2() = (98)

in order to represent the straight line.

N

()

1

1 N2

xx =2

(x)

1

2

3

j x =4j+1 =0

1

2

3

=1

()

3

2

=0 =1

a) b) c)

Figure 9: (a) straight line in the x coordinate space. (b) line after the transformationinto the coordinate space. (c) representation of this straight line by usingthe shape functions N1 and N2 in a linear combination (see expression 99)

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By using these two shape functions the straight line can be represented as a linearcombination of these two shape functions as shown in figure (9).

() = N1 (xj) + N2 (xj+1)

= (1 ) (xj) + (xj+1) (99)

= (1 ) uj + uj+1

where uj and uj+1 are the values of the function u and the approximation uh at thenodes j and j + 1. Inserting the given values from figure (9) for (xj) and (xj+1) wecan write for expression (99)

() = (1 ) 2 + 3

( = 0) = 2

( = 1) = 3

The linear approximation in x coordinates for the given values is then

(x) = 1 + 0, 5x (100)

and in coordinates

() = 2 + (101)

0 1

u(x)

4 u2

x

1

2

3

u1

x =2 x =5 x =8 x =12432

u3

u4

uh(x)

e1 e3e2node1 node4node3node2

u(x)

Figure 10: Subdivision of a bar into three elements with two nodes each. The exact so-lution u(x) is approximated with linear segments. The values of the unknowncoefficients i are the values of the function u(x) = uh(x) at the nodes. Thispreserves the continuity of the piecewise linear approximation uh(x).

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4.1.2. Transformation of the derivatives for a linear 1-D element

For the example in the next chapter we also need the derivatives of the transformationformula (95) and some definitions:

x = xj + (xj+1

xj)

dx

d= xj+1 xj

:= Lj

dx = Ljd (102)

where Lj is the length of the subarea (the finite element). With the chain rule forderivatives we receive:

j[(x)] =

d(x)

dx

dj [(x)]

d

= L1jdj

d(103)

4.2. 1-D Example

Let us now consider the following differential equation of second order

d

dxh(x)

du

dx+ f(x) = 0 (104)

where f(x) is the function for the body forces and the h(x) describes the spatial variationof the material property (e.g. the Young Modulus E). The differential equation is definedwithin the interval

x [a, b]

u(a) = ua and u(b) = ub

According to the variation principal we receive the functional

I(u) = ba

12

h(x)dudx2f(x)u dx . (105)

Let us now subdivide the area in subareas

a = x1 < x2 < ... < xm+1 = b

Dj := [xj, xj+1]; j = 1,...,m

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and uh(x) the approximation of the wanted solution u(x) Expression (105) can now bestated as the sum over the subinterval Dj

I(uh) =m

j=1 Dj1

2h(x)

duh

dx 2

f(x)uh

dx (106)

and for each subinterval Dj we can then write

Ij(uh) =

xj+1xj

1

2h(x)

duh

dx

2f(x)uh

dx (107)

Mapping of the approximation function

uh(x) = (x) = 0 + 1x (108)

onto a standard interval [0, 1] in the coordinates space and inserting the approximation

uh() = () = 0 + 1 (109)

into expression (107) by using expressions (102) and (103) gives

Ij(i) =

10

1

2h()

L1j

dj

d

2f()j

Ljd

=1

2Lj

10

h()

dj

d

2d Lj

10

f()jd (110)

Assuming that the intervals Dj are small enough the functions f(x) and h(x) is constantwithin the interval. Inserting the abbreviations

f =1

2[f(xj) + f(xj+1)] (111)

h =1

2[h(xj) + h(xj+1)] (112)

into expression (110) we receive

Ij(i) =h

2Lj

10

dj

d

2d f Lj

10

jd

= h2Lj

21 f Lj12

(20 + 1) (113)

with 10

dj

d

2d =

10

21d

= 21 (114)

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and 10

jd =

10

(0 + 1)d

=

1

2(20 + 1) . (115)

When uj is the value of the continuous function uh at the node j with its coordinatesxj with j = 1, 2,....,m the following relations are valid:

uj = uh(xj)

= j(xj)

= j( = 0)

= 0 (116)

uj+1 = uh(xj+1) = j(xj+1)

= j( = 1)

= 0 + 1 (117)

Stating expressions (116) and (117) in matrix gives01

=

1 0

1 1

uj

uj+1

:= p u(j) (118)

For 21 from expression (114) the matrix form is

21 =

01

T0 00 1

01

= u(j)T

pT

0 00 1

p

u(j)

= u(j)T

1 1

1 1

u(j)

:= u(j)T

K u(j)

(119)

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where K is the so called the stiffness matrix since here the physical properties are stored.For 0.5(20 + 1) from expression (115) the matrix form is

1

2(20 + 1) =

01

T1

2

21

(120)

= u(j)T

pT12

21

= u(j)

T

1 10 1

1

2

21

= u(j)T

0.50.5

:= u(j)

T

b

where b is defined as

b := 0.50.5 (121)With the following two definitions

j :=h

Lj(122)

j := f Lj (123)

we can express equations (113) as

Ij = u(j)T 1

2j Ku(j)

u(j)T(j b) (124)

Looking for a minimum of Ij the first variation leads to11

Ij =Ij

u(j)= 0 (125)

j (Ku(j)) j b = 0 (126)

As an example we assume that we have four elements. In matrix form we receive thefollowing equations for each element12.

e1 : 1

1 1 0 0 01 1 0 0 0

0 0 0 0 00 0 0 0 00 0 0 0 0

u1u2u3u4u5

= 12 11000

(127)11The unknown variables are now the values ofuj at the nodes (nodal values). Therefore the variation

has to be made towards uj .12In order to be able to combine the single equations into a big set of equations the vectors u(j) are

filled up to a vector u with the missing u1 uj1, uj+2 um+1and the matrixes K are filled withzeros in order to bring to the correct dimension of the problem which is a (m + 1 m + 1)-matrix.

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e2 : 2

0 0 0 0 00 1 1 0 00 1 1 0 00 0 0 0 00 0 0 0 0

u1u2u3u4u5

= 12

02200

(128)

e3 : 3

0 0 0 0 00 0 0 0 00 0 1 1 00 0 1 1 00 0 0 0 0

u1u2u3u4u5

= 12

00330

(129)

e4 : 40 0 0 0 00 0 0 0 0

0 0 0 0 00 0 0 1 10 0 0 1 1

u1u2

u3u4u5

= 12 00

044

(130)

Combining these four expressions for the four elements through superposition we receivefor the global stiffness matrix K(g)

K(g) =

1 1 01 1 + 2 2

2 2 + 3 33 3 + 4 4

0 4 4

(131)

and

b(g) =1

2(1, 1 + 2, 2 + 3, 3 + 4, 4)

T (132)

u(g) = (u1, u2, u3, u4, u5)T (133)

This leads to the following set of equations.

K(g)u(

g) = b(

g) (134)

The difference to the Ritz approach to the problem is obvious. The global stiffnessmatrix K(g), which is the coefficient matrix, has now a so called bandwidth structure,i.e. only in the diagonal of the matrix and some parallel lines we have numbers. Mostof the matrix consists of zeros which makes it very easy to invert the matrix and solvethe problem. The term stiffness matrix expresses that in this matrix the physicalbehaviour of the model is stored.

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5. Finite Element Method - Engineering approach

5.1. Engineering approach for the 1-D example

Another possibility to approach the Finite Element Method is to derive the equations

received in the previous chapter by looking at an easy example. This is again a one-dimensional bar which is elongated along the x axis as shown in figure (11). This bar isdivided into three segments of the length L1, L2, L3 (Finite Elements) with three differentYoungs Modules E1, E2, E3.

x

=10 MPa

0 4

E

0

1

2 31

E2 E3

L 1 L 2 L 3

Figure 11: The bar is fixed on the left hand side at the wall. It is divided into threeelements e1, e2, e3 of the length L1, L2, L3. Material properties within theelements are the Youngs Modules E1, E2, E3. The bar is pulled by a surfaceforce of 0 = 10M P a at its right side.

Figure (12) shows the Element e2 with its displacements u2 and surface force F2 acting

on its left side and the displacements u3 and surface force F3 acting on its right side.

u2

E2

F2

L2

F3

u3

E2

Figure 12: Sketch of the element e2 with the length L2 with the surface forces F2, F3 anddisplacements u2, u3 at its left and right hand side.E2 is the Youngs modulus.

The Hookes law in each individual Finite Element can be written as

= Ei (135)

=du

dx=

ui ui+1

Li

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where Li is the length of each element. According to Hookes law the surface force shownin figure (11) acting on the left-hand-side and the right-hand-side respectively can bewritten as

F2 = 2 =E2(u2 u3)

L2(136)

F3 = 3 =E2(u3 u2)

L2(137)

These two expressions can also be stated in matrix form and we receive

E2

L2

1 1

1 1

u2u3

=

F2F3

K(2)u(2) = b(2) (138)

K(2) = k(2)11 k(2)12k(2)21 k(2)22 =E2

L2 1 1

1 1 (139)

For the elements e1 and e2 we can write in the same way

E1

L1

1 1

1 1

u1u2

=

F1F2

K(1)u(1) = b(1) (140)

K(1) =

k(1)11 k

(1)12

k(1)21 k

(1)22

=

E1

L1

1 1

1 1

(141)

E3

L3

1 1

1 1

u3u4

=

F3F4

K(3)u(3) = b(3) (142)

K(3) =

k(3)11 k

(3)12

k(3)21 k

(3)22

=

E3

L3

1 1

1 1

(143)

where for each element K(e) is the symmetric, quadratic element stiffness matrix, u(e)

the (displacement) vector of the nodal value (or in short node vector) and b(e)

the vectorof the acting surface forces. By looking at node i which connects Element ei and ei+1we receive according to the equilibrium of the surface forces:

Fi F(e)i F

(e+1)i = 0 (144)

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Fi

Fi

(e) Fi

(e+1)

ei ei+1

Figure 13: Sketch of the node i which connects the elements ei and ei+1. The surface

force Fi at the node i has to be in equilibrium with the surface forces F(e)i

and F(e+1)i .

For the four nodes we then receive four equations

F(1)1 = F1 k

(1)11 u1 k

(1)12 u2 = F1 (145)

F(1)2 + F

(2)2 = F2 k

(1)21 u1 + k

(1)22 u2 + k

(2)11 u2 k

(2)12 u3 = F2 (146)

F(2)3 + F

(3)3 = F3 k

(2)21 u2 + k

(2)22 u3 + k

(3)11 u3 k

(3)12 u4 = F3 (147)

F(3)4 = F4 k

(3)21 u3 k

(3)22 u42 = F4 (148)

These equations can be summarized in a global assembly and we receive the followingglobal Finite Element equation in matrix form

k(1)11 k

(1)12 0 0

k(1)21 k

(2)22 + k

(2)11 k

(2)12 0

0 k(2)21 k

(2)22 + k

(3)11 k

(3)12

0 0 k(3)21 k(3)22

u1u2u3u4

=

F1F2F3F4

(149)

This can be written as

K(g) u(g) = b(g) (150)

where K(g) is the global stiffness matrix, u(g) the global displacement vector and b(g)

the global load vector.

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6. Special topics on the Finite Element Method

6.1. Do we need shape functions?

The term shape function is used for a different description of the approximation func-

tion for an element. But it is not compulsory to use shape functions explicitly in orderto define a Finite Element. This is shown in the example in chapter 4 .2 where no shapefunction was used for the development of the Finite Element equation.

However, in more complicated elements of higher order in 3-D it is much more conve-nient to integrate the equations when the approximation function within the element isdescribed as a linear combination of the shape functions. The properties of the shapefunction are:

1. They can have the values one and zero.2. They are only defined within the element. Outside the element their value is zero.

6.2. More definitions of Finite Elements6.2.1. 1D Finite Element with a quadratic polynomial

By choosing a quadratic polynomial for the approximation function uh(x) we receive athird node, i.e. a third unknown coefficient, for the 1D Finite Element.

0 10.5

node 1 node 2 node 3

Figure 14: 1-D element with a quadratic polynomial, i.e. an element with three nodes.

() = 0 + 1 + 22 (151)

From figure (14) we can express the values of the function uh(x) at the nodal points as

u1 = uh(x1) = j(x1) = 1( = 0) (152)

= 0

u2 = uh(x2) = j(x2) = 2( = 0.5)

= 0 + 0.51 + 0.252u3 = uh(x3) = j(x3) = 3( = 1)

= 0 + 1 + 2

Solving this for the i and writing the three expressions in matrix form we receive 01

2

=

1 0 03 4 1

2 4 2

u1u2

u3

(153)

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0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10. 2

0

0.2

0.4

0.6

0.8

1

1.2

N1

N2

N3

Figure 15: Graphs of the three shape functions for a 1-D element with a quadratic poly-nomial

The three shape functions are

N1() = 1 3 + 22 (154)

N2() = 4 42 (155)

N3() = + 22 (156)

and uh() can the be expressed as a linear combination of the shape functions

uh() = N1 u1 + N2 u2 + N3 u3 (157)

The three shape function N1, N2, N3 are drawn in figure (15).

6.2.2. Linear transformation for Triangle Elements

Before we can have a look at the definitions for the triangle element, we have to statethe mapping function for an arbitrary triangle Ti in the xi coordinate system with thenodal points P1(x

(1)1 , x

(1)2 ), P2(x

(2)1 , x

(2)2 ) and P3(x

(3)1 , x

(3)2 ) onto a standard unit triangle

T0 in the i coordinate system (see figure 16).The transformation formula for the mapping from the xi coordinate system into the

i coordinate system is

x1 = x(1)1 + [x

(2)1 x

(1)1 ]1 + (x

(3)1 x

(1)1 )2 (158)

x2 = x(1)2 + [x

(2)2 x

(1)2 ]1 + (x

(3)2 x

(1)2 )2 (159)

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P0 1

1

1

P2

P3

1

2

x

P

0

1

P2

P3

1

2x

Figure 16: Mapping of an arbitrary triangle Ti in the xi coordinate system with thenodal points P1(x

(1)1 , x

(1)2 ), P2(x

(2)1 , x

(2)2 ) and P3(x

(3)1 , x

(3)2 ) onto a standard unit

triangle T0 in the i coordinate system.

6.2.3. 2-D triangular elements with a linear polynomial

By choosing a linear polynomial for the approximation function uh(x) we receive for the2D Triangular Element three nodes, i.e. three unknown coefficients i (see figure 16):

() = 0 + 11 + 22 (160)

Again we can express the values of the function uh(x) at the nodal points as

u1 = uh(x1i ) = 1(1 = 0, 2 = 0) (161)

= 0

u2 = uh(x2i ) = 2(1 = 1, 2 = 0)

= 0 + 1u3 = uh(x

3i ) = 3(1 = 0, 2 = 1)

= 0 + 2

Solving this for the i and writing the three expression in matrix form we receive 01

2

=

1 0 01 1 0

1 0 1

u1u2

u3

(162)

The three shape functions are

N1(1, 2) = 1 1 2 (163)

N2(1, 2) = 1 (164)

N3(1, 2) = 2 (165)

and uh() can be expressed as a linear combination of the shape functions

uh(1, 2) = N1 u1 + N2 u2 + N3 u3 (166)

Shape function N1 is drawn as an example in figure (17).

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P0 1

1

1

P2

P3

1

2

( )i

Figure 17: Shape function N1 for a Triangular Element with a linear polynomial.

6.3. Classification of the various element types

According to space

1-D e.g. line, bar, cantilever2-D e.g. triangle, quad3-D e.g. Tetraeder, Hexaeder

According to the degree of the polynomial in 1-D

0 + 1x (linear)0 + 1x + 2x

2 (quadratic)0 + 1x + 2x

2 + 3x3 (cubic)

(167)

According to completeness of the polynomial and properties

Isoparametric: all elements where their transformation functionpreserves the geometrical shape

Lagrange: complete polynomial

Serendipity: incomplete polynomial

Conform: every element which fulfills continuity ofthe node variables at the nodes

For example, if a quad

element with a complete bi-quadratic polynomial is given it belongs to the Lagrangeclass. A quad element with a bi-quadratic polynomial belongs to the Serendipity class.

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A. Mathematical rules and equations

A.1. Positive definite differential operator

Under certain conditions there is a relationship between a given differential equation

and a variational formulation. Mathematically spoken: If a linear differential opera-tor is positive definite, the differential equation can always be stated as a variationalformulation. If for the inner product

< v, w > =

V

v w dV (168)

the following equations are true

< L v,w > = < v, L w > L is self-adjoint, i.e. symmetric

< L v, v > > 0 L is elliptic (169)

the differential operator L is positive definite (e.g., the differential operator Laplace and the differential operator Nabla operator are positive definite).

A.2. Calculus rules for the first variation

The calculus rules for the first variation are

(u + v) = u + v

(u) = (u)

(u v) = v u + u v (170)

Example from equation (35) in order to receive equation (36)

u

x

2=

u

x

u

x

=

u

x

u

x+

u

x

u

x

= 2

u

x

u

x

= 2

u

x

x(u)

A.3. Integration by parts

By integrating the product rule of differentiation one gets the integration by partsV

(u v) =

V

u v +

V

v u

V

u v =

V

(u v)

V

u v (171)

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The strike at the brackets in (u v) denotes the spatial derivation. Example in order toreceive equation (37)

V

a1u

x1

x1(u)

dV (172)

u = a1u

x1and v =

x1(u) (173)

u =

x1

a1

u

x1

v = u

V

a1

u

x1

x1(u)

dV = (174)

V

x1

a1

u

x1u

dV V

x1

a1

u

x1

u

dV (175)

A.4. Gauss integral equation

The integral sentence of Gauss (divergence of a vector field) converts a volume integralto a closed surface integral.

V

div u dV =

S

(u n) dS (176)

A.5. Stokes integral equation

The integral sentence of Stokes (rotation of a vector field) converts a surface integralinto a closed line integral integral.

S

(rotu n) dS =

C

u dC (177)

For a plane case we receive

rotu n =u2

x1

u1

x2. (178)

A.6. Stress definitions

The stress tensor can be transformed into a diagonal matrix. The remaining diagonalcomponents are the principal stresses. 1, 2 and 3. For these it is defined that

1 > 2 > 3 . (179)

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In Geoscience compressive stresses are positive. In engineering it is just the other wayaround!

The three invariants of the stress tensor expressed with the principal stresses are

I1 = 1 + 2 + 3

I2 = (23 + 31 + 12)I3 = 123 (180)

The mean stress (pressure) is

m = p =1

3(1 + 2 + 3) (181)

The differential stress is defined as

d = 1 3 (182)

and the deviatoric stress as

ij = ij m (183)

In the presence of pore fluids, the pore fluid pressure f, reduces the normal stresses.The effective normal stress is defined as

neff = n f (184)

The brittle or plastic strength y is a threshold value above which the rock material can

not resist the stresses. It breaks (yields).

ymax =1

2(1 3) = C + n (185)

where is the friction and C the cohesion. Both values are different depending if theyield strength is prescribed for existing faults (Amontons law) or newly created fractures(Mohr-Coulomb criteria). For the friction on existing faults Byerlee (1966) determinedC and in the laboratory with

ymax =1

2(1 3) = 0.85n (186)

when pressure is below 200 MPa (ca. 8 km depth) and

ymax =1

2(1 3) = 60M P a + 06n (187)

for deeper layers. In the literature this is stated as the Byerlee law.

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A.7. Strain tensor

The strain tensor describes the deformation (Volume change and shape change) of abody.

ij =1

2ui

xj +uj

xi (188)

After transformation into the principal system the three remaining diagonal componentsare the principal strains 1, 2 and 3 and it is defined

1 > 2 > 3 (189)

The three invariants of the strain tensor expressed in the principal strains are

E1 = 1 + 2 + 3

E2 = (23 + 31 + 12)E3 = 123 (190)

x2

0

P

P'

Q'

Q

D'

D

ui+dui

ui

x1

x3

Figure 18: Displacement of two points P and Q of a body D towards the points P andQ of the new shape of the body D.

In order to deduce the deformation tensor and extraction of the non deforming parts

of translation and rotation two points P and Q according figure (18) of the body Dare considered. After deformation towards P and Q the body has the new shape andposition D. The displacement of the two points P and Q is expressed through thedisplacement vectors ui and ui + dui. In order to express ui(x) at an arbitrary location,

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ui(x) in written in a Tayler expansion at point P.

ui(Q) =n=0

(Q P)nuni (P)

n!

= ui(P) + uixjPxi= ui(P) +

1

2

ui

xj+

uj

xi

uj

xi+

ui

xj

xi

= ui(P) +1

2

ui

xj

uj

xi

xi +

1

2

ui

xj+

uj

xi

xi (191)

The first term is the translation, the second the describes the antisymmetric rotationtensor and the third is the deformation tensor. The symmetric deformation tensor canbe split into an isotropic part which is a pure volume change and shear part which

reflects the shape change of the body.

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B. List of useful books and papers

B.1. Books and paper quoted in the text

Altenbach, J. and Sacharov, A.S., 1982. Die Methode der Finiten Elemente in der

Festkorpermechanik. VEB Fachbuchverlag, Leipzig.

Bird, P. and Kong, X., 1994. Computer simulations of California tectonics confirmvery low strength of major faults. Geol. Soc. of Am. Bull., 106: 159-174.

DeMets, C., Gordon, R.G., Argus, D.F. and Stein, S., 1990. Current Plate Mo-tions. Geophys. J. Int., 101: 425-478.

DeMets, C., Gordon, R.G., Argus, D.F. and Stein, S., 1994. Effect of recent revi-sions to the geomagnetic reversal time scale on estimates of current plate motions.Geophys. Res. Lett., 21/20: 2191-2194.

Kammel, G., Franeck, H. and Recke, H.-G., 1990. Einfuhrung in die Methode derfiniten Elemente. Carl Hanser Verlag, Munchen.

Ranalli, G., 1995. Rheology of the Earth. Chapman and Hall, London.

Schwarz, H.R., 1991. Methode der finiten Elemente. Teubner, Stuttgart.

Sperner, B., F. Lorenz, et al. (2001). Slab break-off - abrupt cut or gradualdetachment? New insights form the Vrancea Region (SE Carpathians, Romania).Terra Nova 13: 172-179.

Stuwe, K., Geodynamik der Lithosphare, Springer, Berlin, 2000. Turcotte, D.L., and G. Schubert, Geodynamics, Cambridge University Press, Cam-

bridge, 2002.

Zienkiewicz, O.C. and Tayler, R.L., 1994a. The Finite Element Method, Volume 1:Basic Formulations and Linear Problems, Fourth Edition. McGraw Hill, London.

Zienkiewicz, O.C. and Tayler, R.L., 1994b. The Finite Element Method, Volume 2:Solid and Fluid Mechanics, Dynamics and Non-linearity, Fourth Edition. McGrawHill, London.

B.2. Books on FEM for further reading

Bathe, K.J., 1982. Finite Elemente Methode. Springer Verlag, Berlin.

Braess, D., 1992. Finite Elemente. Springer Verlag, Berlin.

Braun, J. and Sambridge, M., 1995. A numerical method for solving partial dif-ferential equations on highly irregular evolving grids. Nature, 376: 655-660.

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Clough, R.W., 1960. The Finite Element in Plane Stress Analysis, Proc. 2nd.A.S.C.E. Conf. on Electronic Computation, Pitsburgh, pp. 114-123.

Courant, R., 1943. Variational Methods for the Solution of Problems of Equilib-rium and Vibrations. Bull. Am. Math. Soc., 49.

Cuvelier, C., Segal, A. and van Steenhoven, A.A., 1986. Finite Element Methodsand Navier-Stokes Equations. D. Reidel Publishing Company, Dordrecht, Holland.

Meissner, U. and Menzel, A., 1989. Die Methode der finiten Elemente. SpringerVerlag, Berlin.

Nasitta, K. and Hagel, H., 1992. Finite Elemente. Mechanik, Physik und nicht-lineare Prozesse. Springer Verlag, Berlin.

Ritz, W., 1909. Uber eine neue Methode zur Losung gewisser Variationsproblemeder mathematischen Physik. J. Reine Angew. Math., 135.

Tornig, W., 1985. Numerische Losung von partiellen Differentialgleichungen derTechnik. Teubner, Stuttgard.

van Kan, J.J.I.M. and Segal, A., 1995. Numerik partieller Differentialgleichungenfur Ingenieure. Teubner, Stuttgart.

B.3. Papers on FE-model in Geoscience

Albarello, D., Mantovani, E. and Viti, M., 1997. Finite Element Modeling of therecent-present deformation pattern in the Calabrian arc and surrounding regions.

Annali Geofisica.

Bassi, G., Sabadini, R. and Rebai, S., 1997. Modern tectonic regime in the Tyrrhe-nian area: observations and models. Geophys. J. Int., 129: 330-346.

Becker, T.W., Faccenna, C., OConnell, R.J. and Giardini, D., 1999. The de-velopment of slabs in the upper mantle: insights from numerical and laboratoryexperiments. J. Geophys. Res., 104: 15207-15226.

Bird, P. and Piper, K., 1980. Plane-stress finite-element models of tectonic flow insouthern California. Phys. Earth Planet. Int., 21: 158-175.

Bird, P., 1998. Testing hypotheses on plate-driving mechanisms with global litho-sphere models including topography, thermal structure, and faults. J. Geophys.Res., 103: 10115-10129.

Bott, M.H.P., 1990. Stress distribution and plate boundary force associated withcollision mountain ranges. Tectonophysics, 182: 193-209.

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Bott, M.P.H., 1991. Ridge push and associated plate interior stress in normal andhot spot regions. Tectonophysics, 200: 17-32.

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Finite Elementet Lecture Fluid (Good)

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