Introduction into Finite Elementsta.twi.tudelft.nl/users/vermolen/wi3098/wi3098.pdf · The notes provide an introduction into the Finite Element method applied to Partial Diﬀerential

Introduction into Finite Elements

Fred Vermolen and Domenico Lahaye

Delft University of Technology

Delft Institute for Applied Mathematics

Section of Numerical Analysis

Version 2.1

February 26, 2008

Preface

These lecture notes are used for the course Finite Elements (wi3098LR) forBSc-students of the department of aerospace engineering with a minor in ap-plied mathematics, and for students visiting the Delft University of Technologyin the framework of ATHENS. For the interested student some references aregiven for further reading, but note that the literature survey is far from com-plete.

The notes provide an introduction into the Finite Element method applied toPartial Differential Equations. The treated problems are classical. The treat-ment of the Finite Element Method is mathematical, but without the conceptof Hilbert and Sobolev spaces, which are fundamental in a rigorous treatmentof the Finite Element Method. Recently, a more detailed and comprehensivetreatment appeared by van Kan et al. [2006], which is available from the DelftUniversity Press.

Further, we recommend all students of this course to attend the lectures sincethe lecture notes do not aim at being complete.

We wish everybody good luck with the material and the course! Further, wewould like to thank Caroline van der Lee for the beautiful typesetting of thisdocument. Finally, we thank Fons Daalderop and Martin van Gijzen for theircritical reading of these lecture notes and valuable suggestions to improve it.

- Alfonzo e Martino, grazie!

Fred Vermolen and Domenico LahayeDelft Institute of Applied Mathematics

Delft University of [email protected]

c©Copyright Fred J. Vermolen, 2008

Contents

Preface i

1 Model Equations 11.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Mathematical Preliminaries . . . . . . . . . . . . . . . . . . . . . 5

2 Minimization Problems 92.1 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92.2 A general 1-dimensional minimization problem . . . . . . . . . . 132.3 A 2-dimensional minimization problem . . . . . . . . . . . . . . . 152.4 Some formalization . . . . . . . . . . . . . . . . . . . . . . . . . . 17

3 Variational Formulation and Differential equations 203.1 Weak forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203.2 Which weak formulation? . . . . . . . . . . . . . . . . . . . . . . 243.3 Mathematical considerations:existence and uniqueness . . . . . . 25

4 Galerkin’s Finite Element method 284.1 The principle of Galerkin’s method . . . . . . . . . . . . . . . . . 284.2 Motivation of piecewise linear basis-functions . . . . . . . . . . . 314.3 Evaluation of a one-dimensional example . . . . . . . . . . . . . . 324.4 Ritz’ method of Finite Elements by a simple example . . . . . . . 344.5 The treatment of a non-homogeneous Dirichlet BC . . . . . . . . 354.6 A time-dependent example . . . . . . . . . . . . . . . . . . . . . 374.7 The principle of element matrices and vectors . . . . . . . . . . . 384.8 Numerical integration . . . . . . . . . . . . . . . . . . . . . . . . 424.9 Error considerations . . . . . . . . . . . . . . . . . . . . . . . . . 43

5 Time dependent problems: numerical methods 465.1 Time-integration methods . . . . . . . . . . . . . . . . . . . . . . 465.2 Accuracy of Time-integration methods . . . . . . . . . . . . . . . 475.3 Time-integration of PDE’s . . . . . . . . . . . . . . . . . . . . . . 49

CONTENTS iii

5.3.1 The Heat equation . . . . . . . . . . . . . . . . . . . . . . 495.3.2 The wave equation . . . . . . . . . . . . . . . . . . . . . . 51

5.4 Stability analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

6 Numerical linear algebra 556.1 Basics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

6.1.1 Eigenvalues and eigenvectors of A . . . . . . . . . . . . . 566.1.2 Diagonalization of A . . . . . . . . . . . . . . . . . . . . . 576.1.3 Eigenvalues of a symmetric matrix . . . . . . . . . . . . . 57

6.2 Solution of systems of linear equations . . . . . . . . . . . . . . . 586.2.1 The LU -decomposition . . . . . . . . . . . . . . . . . . . 586.2.2 Basic, classical iterative solvers . . . . . . . . . . . . . . . 62

6.3 Convergence properties of iterative methods . . . . . . . . . . . . 636.4 Gradient methods for Ax = b . . . . . . . . . . . . . . . . . . . . 68

Bibliography 71

1

Model Equations

1.1 Introduction

In technology and engineering there is much to model and predict. The pre-dictions are often based on mathematical models. The models include bothphysical/chemical assumptions and mathematical computation techniques. Theanswers from the models should be reliable, hence accuracy, consistency andstability are important issues in the analysis of the models.

Since mathematical models can become complicated, numerical (approxi-mation) techniques are often necessary to find a prediction. The numericaltechniques give unrealistic predictions when they are not used properly.

This course aims at:

• Interpretation of numerical results;

• Analysis of numerical methods;

• Analysis of mathematical models to get a qualitative view of the predic-tion without actually computing it.

This course will include items of:

• Linear Algebra (matrix computation),

• Numerical analysis,

• Calculus,

• Differential equations.

Before we sum the partial differential equations, which will be the basis ofthe course, we introduce some notation:

1.1 Introduction 2

[1] A position vector in a two-dimensional plane and three-dimensional spaceis denoted as a row of respectively two and three numbers a1, a2 and a3

separated by a comma, i.e.

a =< a1, a2 > in 2-D, or a =< a1, a2, a3 > in 3D.

The overbar is used to indicate that we mean a position vector. Thecoefficients a1, a2 and a3 represent the x, y and z components of thevector.

[2] A vector in the linear algebra sense can have a length n, where n is apositive integer, is denoted by a column, i.e.

b =

b1

b2

b3

. . .

bn

.

The underbar is used to indicate that we mean a vector in the linearalgebra sense. The vector may represent a result as the solution of alinear system of equations. This will be used in the later chapters of thenotes.

[3] We introduce the gradient of a scalar function as the vector containingthe directional derivatives with respect to the co-ordinate axes, i.e. for afunction f = f(x, y) and a function g = g(x, y, z):

∇f =<∂f

∂x,∂f

∂y> and ∇g =<

∂g

∂x,∂g

∂y,∂g

∂z> .

[4] The laplacian is defined as follows:

∆(·) := ∇ · ∇(·)

=

⟨

∂

∂x,

∂

∂y,

∂

∂z

⟩

·

⟨

∂

∂x,

∂

∂y,

∂

∂z

⟩

(·)

=∂2

∂x2(·) +

∂2

∂y2(·) +

∂2

∂z2(·). (1.1)

This is the sum of the second derivatives of a function with respect to allthe co-ordinate axes.

We will consider the following PDE’s (Partial Differential Equations) as abasis for many mathematical models:

1.1 Introduction 3

− ∆u = f Poisson, Laplace (f = 0) (1.2)

− ∇ · (D(u)∇u) = f non-linear Poisson (1.3)

∂u

∂t= ∇ · (D(u)∇u) diffusion (1.4)

∂u

∂t+

∂f(u)

∂x= 0 Buckley-Leveret / two-phase flow (1.5)

∂u

∂t+ ∇ · (qu) = ∆u convection-diffusion (1.6)

∂2u

∂t2= c2∆u wave propagation (1.7)

The above PDE’s originate from various physical or biological models. Inorder to have a unique (well defined) solution to the PDE, we need boundaryand initial conditions. We will treat the most common boundary conditionsbriefly. Let us assume that u satisfies a PDE within the domain of computationΩ.

[1] The first boundary condition concerns the situation in which the solutionis given at the boundary, that is

u = g1(x), for x ∈ ∂ΩD. (1.8)

The solution is well defined at ΩD which represents (part of) the boundaryof the domain Ω. This boundary condition is referred to as a Dirichletboundary condition. Physically, this could correspond to a prescription ofthe temperature at that (part of the) boundary.

[2] The second boundary condition concerns the case in which the normalderivative is prescribed, that is

∂u

∂n= g2(x), for x ∈ ∂ΩN . (1.9)

This boundary condition is referred to as a Neumann boundary condition.Physically, this could correspond to a prescription of the flux of matteror heat on this (part of the) boundary. This condition is also referred toas a flux boundary condition.

[3] The third boundary condition concerns the case in which a linear combi-nation of the solution and its normal derivative is specified, that is

σu +∂u

∂n= g3(x), x ∈ ΩR. (1.10)

Here σ is a nonnegative real number. This boundary condition is referredto as a Robin boundary condition. People also call it a mixed boundarycondition. Physically, it may correspond to radiation.

1.2 Mathematical Preliminaries 4

The time dependent problems also need one initial condition if the maximumorder of the time derivative is one and more initial initial conditions of themaximum order of the time derivative is larger than one. We will discuss theboundary and initial conditions for three standard PDE’s.

Poisson’s equation

Consider Poisson’s equation−∆u = f(x), (1.11)

where f(x) is a given function. For this PDE, we have precisely one boundarycondition on each point on the boundary. In the mathematical literature, thisequation is referred to as elliptic.

The heat equation

Consider the heat equation

∂u

∂t− ∆u = f(x), (1.12)

then we need one boundary condition on each point on the boundary like in theprevious example. Furthermore, due to the time derivative, we need an initialcondition to hold at each point of the domain of computation Ω:

u(x, t) = u0(x), for x ∈ Ω. (1.13)

Here u0(x) is a known function. In the mathematical literature, the aboveequation is referred to as parabolic.

The wave equation

Consider the wave equation

∂2u

∂t2− ∆u = f(x), (1.14)

then we need one boundary condition on each point on the boundary like in theprevious example. Furthermore, due to the second order time derivative, weneed two initial conditions to hold at each point of the domain of computationΩ:

u(x, t) = u0(x), for x ∈ Ω,∂u

∂t(x, t) = v0(x), for x ∈ Ω.

(1.15)

Here u0(x) and v0(x) are known functions. In the mathematical literature, theabove equation is referred to as hyperbolic.

The classification of the PDE’s into hyperbolic, parabolic and elliptic only ap-plies for second order PDE’s and has been inspired by quadratic curves. Theclassification is not treated in more detail.


∂Ωn

Ω

Figure 1.1: A domain of computation Ω with its boundary ∂Ω and outward unit normalvector n

SΓ

n

Figure 1.2: A surface S in R3 with boundary curve Γ and unit outward normal vectorn

1.2 Mathematical Preliminaries

Let Ω ⊂ R3, ∂Ω = Ω\Ω (the boundary of Ω), and n is the outward unit normalon ∂Ω with ‖n‖ = 1 (figure 1.1). Further Ω is regular and closed. Then:

[1] Divergence theorem of Gauss.

Let Ω ⊂ R3 be a regular closed body with outward unit normal vector n(‖n‖ = 1) on boundary ∂Ω, and let F (x, y, z) be a differentiable vector-function on Ω, then

∫∫∫

Ω

∇ · F (x, y, z)dV =

∫∫

∂Ω

F (x, y, z) · ndS, (1.16)

or shortly∫

Ω

∇ · F dV =

∫

∂Ω

F · ndS. (1.17)

[2] Circulation theorem of Stokes.

Let S ⊂ R3 be a closed regular oriented surface bounded by curve Γ with


Ω

Γ

Figure 1.3: A domain Ω in R2 with boundary Γ

positive orientation (figure 1.2), then

∫∫

S

(∇ × F (x, y, z)) · ndS =

∫

Γ

F (x, y, z) · dr, (1.18)

or shortly∫

S

curlF · ndS =

∫

Γ

F · dr. (1.19)

here n is the unit normal and F is assumed to be differentiable over S.

[3] Divergence theorem for R2

Let Ω ⊂ R2 be a regular closed area in R2 with outward unit normal vectorn (‖n‖ = 1) on boundary Γ (figure 1.3), and let F (x, y) be a differentiablevector-function on Ω, then

∫

Ω

∇ · F (x, y)dA =

∮

Γ

F (x, y) · nds. (1.20)

Proof of the above theorems can be found in: Adams [1999], Steward [2003],Almering e.a. [1987], . . .

Examples:

[1]

F (x, y, z) = < x, y, z >, (1.21)

∇ · F =3, (1.22)

Ω :=(x, y, z) ∈ R3 : x2 + y2 + z2 ≤ 1, (1.23)

∂Ω :=(x, y, z) ∈ R3 : x2 + y2 + z2 = 1, (1.24)

⇒

∫∫∫

Ω

∇ · F dV = 3

∫∫∫

Ω

dV = 3V (Ω) = 34

3π13 = 4π. (1.25)

Furthermore on ∂Ω

n =< x, y, z >

√

x2 + y2 + z2, ∂Ω := (x, y, z) ∈ R : x2 + y2 + z2 = 1, (1.26)


∫∫

∂Ω

F · ndS =

∫∫

∂Ω

< x, y, z > · < x, y, z >1

√

x2 + y2 + z2dS,

=

∫∫

∂Ω

x2 + y2 + z2

√

x2 + y2 + z2dS,

=

∫∫

∂Ω

√

x2 + y2 + z2dS =

∫∫

∂Ω

dS = A(∂Ω) = 4π. (1.27)

[2] Let:

F (x, y, z) = < x, y, z >, (1.28)

∇ × F =

∣

∣

∣

∣

∣

∣

∣

i j k

∂x ∂y ∂z

x y z

∣

∣

∣

∣

∣

∣

∣

= 0, (1.29)

Ω :=(x, y, z) ∈ R3 : x2 + y2 ≤ 1, z = 0, (1.30)

Γ :=(x, y, z) ∈ R3 : x2 + y2 = 1, z = 0, (1.31)

n =(0, 0, 1), (1.32)

⇒

∫∫

Ω

∇× F · ndS = 0. (1.33)

Further on Γ := < x, y, z >∈ R3 : x2 + y2 = 1, z = 0,

∫

Γ

F · dr =

∫

Γ

F (x(t), y(t), z(t)) · dr, (1.34)

x(t) = cos (t), y(t) = sin (t), z(t) = 0, (1.35)

⇒ dr = r′(t)dt =< − sin (t), cos (t), 0 > dt, (1.36)

⇒

∫

Γ

F · dr =

2π∫

0

< cos (t), sin (t), 0 > · < − sin (t), cos (t), 0 > dt = 0.

(1.37)

[3] Consider:

F (x, y) = < x, y >, (1.38)

∇ · F =2, (1.39)

Ω :=(x, y) ∈ R2 : x2 + y2 ≤ 1, (1.40)

Γ :=(x, y) ∈ R2 : x2 + y2 = 1, (1.41)

⇒

∫

Ω

∇ · F dA = 2

∫

Ω

dA = 2A(Ω) = 2π. (1.42)


Furthermore on Γ

n = < x, y >1

√

x2 + y2, (1.43)

∫

Γ

F · ndS =

∫

∂Ω

< x, y > · < x, y >1

√

x2 + y2dS, (1.44)

=

∫

Γ

x2 + y2

√

x2 + y2dS

∫

Γ

√

x2 + y2dS =

∫

Γ

dS = 2π. (1.45)

[4] Example in differential equations. Suppose that

−∆u = f on Ω ⊂ R3,∂u

∂n= g1 on ∂Ω, ∂Ω = Ω \ Ω.

(1.46)

−

∫

Ω

∆udV =

∫

Ω

fdV, (1.47)

−

∫

Ω

∇ · ∇udV = −

∫

∂Ω

n · ∇udS = −

∫

∂Ω

∂u

∂ndS =

∫

Ω

fdV, (1.48)

−

∫

∂Ω

g1dS =

∫

Ω

fdV. (1.49)

The above condition is used as a check whether problem (1.46) has asolution at all. If the above relation (1.49) does not hold, then (1.46) isill-posed (a mathematical word for nonsense). Relation (1.49) can be usedfor a check between f and g.

Exercise 1.2.1. Does (1.50), specified below, have a solution?

−∆u = 1 on Ω := (x, y, z) ∈ R3 : x2 + y2 + z2 < 1,∂u

∂n= 0 on ∂Ω := (x, y, z) ∈ R3 : x2 + y2 + z2 = 1.

(1.50)

Remark. An important question that concerns a mathematical problem is whethera solution for the problem exists (existence), and if so, is the solution thenunique (uniqueness)?

2

Minimization Problems

The application of the Finite Element Method (FEM) is very general. Further,from a mathematical point of view this method is elegant, however also morecomplicated than the method of Finite Differences or Finite Volumes. The mostimportant reasons for the application of finite element methods is the ease of theuse of odd-shaped domains, local mesh refinement, unstructured grids, jumpsin coefficients and use of parallel computing.

The principle of the FEM is explained in two steps

[1] minimization problem,

[2] variational principle.

Ritz’s FEM method is based on the solution of a discrete minimization prob-lem. The method cannot be applied to all PDE’s. The method of Galerkin,based on a variational priciple, can always be applied and, hence, is more gen-eral. The method of Galerkin will be treated in more detail in this course inChapter 4. We will briefly treat some principles of minimization to motivatethe variational principle in the Galerkin’s FEM. We will first see that for somecases minimization problems correspond to PDE’s, i.e. the solution of a mini-mization problem can also be a solution of a PDE with appropriate BoundaryConditions (BC). Later in Chapter 3, we will introduce “weak forms” of PDE’s.After this, we have sufficient background to actually concentrate on the FiniteElement Method to be treated in Chapter 4.

We will see that a minimization problem and a (partial) differential equationsometimes have the same solution under strict smoothness requirements.

2.1 Example

A minimization problem can be (figure 2.1):

2.1 Example 10

y

x

y = f(x)

ba

ya

yb

Figure 2.1: A smooth curve in R2 that connects the points (a, ya) and (b, yb)

Find a smooth curve such that its length between points (a, ya) and (b, yb) isminimal.

In this course a function is smooth if all (partial) derivatives that we needexist. In more advanced courses Sobolev spaces are used to specify the nec-essary smoothness conditions. Let us not be bothered by this in this course.We will formulate this problem in a more mathematical way: the length of thecurve is given by

l(y) =

b∫

a

√

1 + f ′(x)2dx, for any y = f(x). (2.1)

Mathematically

Minimize the integral l(y) over the class of smooth functions which satisfyy(a) = ya and y(b) = yb.

or even more mathematically

Find y, which is continuous and differentiable and satisfies y(a) = ya , y(b) = yb

such thatl(y) ≤ l(y),

for all y smooth and y(a) = ya , y(b) = yb.

Many physical laws are based on minimization problems:

• minimum potential energy,

• Hamilton’s principle in quantummechanics,

• optical length,

• . . .

We go back to example 1:

2.1 Example 11

Find y ∈ Y :=y continuous, differentiable: y(a) = ya, y(b) = ybsuch that l(y) ≤ l(y) ∀y ∈ Y .

The solution y(x) is perturbed with a function v(x), with v(a) = 0 and v(b) = 0,where v(x) is continuous and differentiable. Given any ε ∈ R, with ε a real con-stant, then we put y(x) = y(x) + εv(x). Further

• y(a) = y(a) + εv(a) = y(a) + 0 = ya.

• y(b) = y(b) + εv(b) = y(b) + 0 = yb.

• since y(x) and v(x) are smooth and ε is a constant, it follows that y(x) =y(x) + εv(x) is also smooth.

This implies that y(x) ∈ Y and y(x) ∈ Y . Further we note that we search y(x)such that

l(y(x)) ≤ l(y(x) + εv(x)), (2.2)

∀v(x) smooth and v(a) = 0 = v(b). Further l(y(x) + εv(x)) is minimal forε = 0. Hence d

dεl(y(x) + εv(x))|ε = 0 ∀v ∈ V . Where V := v smooth:

v(a) = 0 = v(b). We will proceed with this:

l(y + εv) =

b∫

a

√

1 + (y′ + εv′)2dx, (2.3)

d

dεl(y + εv) =

d

dε

b∫

a

√

1 + (y′ + εv′)2dx, (2.4)

d

dεl(y + εv) =

b∫

a

d

dε

√

1 + (y′ + εv′)2dx = (2.5)

(

chain rule−−−−−−→

)

=

b∫

a

(y′ + εv′)v′√

1 + (y′ + εv′)2dx. (2.6)

Since ddε

l(y(x) + εv(x))|ε=0 = 0, we obtain

b∫

a

y′v′√

1 + (y′)2dx = 0 ∀v ∈ V. (2.7)

Using partial integration , the above expression is written as:

[y′v

√

1 + (y′)2]ba −

b∫

a

d

dx[

y′√

1 + (y′)2]vdx = 0. (2.8)

Since v(a) = 0 = v(b), the first term in the left-hand side vanishes. Thedifferentiation in the 2nd term of the left-hand side is done by the use of thechain rule:

2.1 Example 12

d

dx

[

y′√

1 + (y′)2

]

=d

dy′

[

y′√

1 + (y′)2

]

· y′′ =y′′

(1 + (y′)2)32

. (2.9)

Hence one obtains:

∫ b

a

y′′

(1 + (y′)2)32

· vdx = 0 ∀v ∈ V. (2.10)

One easily sees that y′′ = 0 on x ∈ (a, b) satisfies the above requirement. Thisimplies that

y′′ = 0 for x ∈ (a, b),

y(a) = ya, y(b) = yb.(2.11)

The minimization is now replaced by a differential equation. Note that werequire y′′ = 0 and that for the differential equation to hold the second deriva-tive must be continuous (and of course exist). Wheras, for the minimizationproblem we only need y′ to exist. The differential equation puts an additionalrequirement on y. Hence the conditions for the solution of the minimizationproblem are weaker than the conditions for the differential equation. Further,we see that y′′ = 0 satisfies (2.10), we did not prove yet that y′′ = 0 is the onlypossibility!

For now we assume that y′′ = 0 is the only possibility, later we will prove thisfact in a lemma, which was proved for the first time by DuBois and Reymond.This can also be found in the books of van Kan et al. [2006] and Strang andFix [1973] for instance.

y′′ = 0 ⇒ y(x) = Ax + B (linear). (2.12)

Combination with the boundary conditions gives

y(x) = ya +yb − ya

b − a(x − a). (2.13)

This is the straight line that minimizes the length of the smooth curve thatconnects (a, ya) and (b, yb). Of course, the result is trivial but the relationbetween a minimization problem and a differential equation is clearly visible.

Now we will establish the Raymond/DuBois lemma:

Lemma 2.1.1. Let F (x) be continuous over [a, b], i.e. F ∈ C0[a, b] and supposethat

b∫

a

F (x)v(x)dx = 0, ∀v ∈ C0[a, b] and v(a) = 0 = v(b)

Then,F (x) = 0 for x ∈ [a, b].

2.2 A general 1-dimensional minimization problem 13

Proof 2.1.1. We use contradiction as an argument. Suppose that F (xo) > 0for any xo ∈ (a, b), note that the case F (xo) < 0 is similar, then since F (x) iscontinuous, there exists a δ > 0 such that

F (x) > 0 whenever |x − xo| < δ.

Now we choose (this is a trick)

v(x) =

(x − xo − δ)2(x − xo + δ)2, |x − xo| < δ,

0, |x − xo| > 0.

Note that v(a) = 0 = v(b) and that v is differentiable and continuous over [a, b].This implies that

b∫

a

F (x)v(x)dx =

x0+δ∫

x0−δ

F (x)(x − xo − δ)2(x − xo + δ)2dx > 0. (2.14)

This violates the hypothesis

b∫

a

F (x)v(x)dx = 0, ∀v.

Hence F (x) > 0 is impossible at any point within [a, b]. Further the case F (x) <0 is dealt with similarly. We conclude that F (x) = 0 on [a, b].

By the use of Lemma 2.1.1 is follows that y′′ = 0 is the only possibility forcondition (relation) (2.10). We will generalize this a little.

2.2 A general 1-dimensional minimization problem

Given the following minimization problem; given f continuous with continuouspartial derivatives

Minimize:

I(u) :=

x1∫

xo

f(x, u, u′)dx,

over all u ∈ C1[x0, x1] with u(x0) = u0.

We will find a differential equation for this minimum u, for which u(x0) = u0.

I(u) ≤ I(u) for all u continuous, differentiable and u(x0) = u0.

Hence ddε

I(u + εv)|ε = 0, with v differentiable and u(x0) + εv(x0) = u0., sov(x0) = 0.

2.2 A general 1-dimensional minimization problem 14

I(u + εv) =

x1∫

xo

f(x, u + εv, u′ + εv′)dx, (2.15)

d

dεI(u + εv) =

x1∫

xo

∂f

∂u(x, u + εv, u′ + εv′)v +

∂f

∂u′(x, u + εv, u′ + εv′)v′

dx.

(2.16)

Furthermore; since ddε

I(u + εv)|ε=0 = 0, we obtain

x1∫

xo

∂f

∂u(x, u, u′)v +

∂f

∂u′(x, u, u′)v′

dx = 0, (2.17)

for all v smooth and v(x0) = 0. Partial integration of the second term gives:

x1∫

xo

∂f

∂uvdx +

[

∂f

∂u′v

]x1

xo

−

x1∫

xo

∂

∂x

(

∂f

∂u′

)

vdx = 0. (2.18)

In other words we have for all smooth functions v(x) smooth and v(x0) = 0:

x1∫

xo

∂f

∂u−

∂

∂x

(

∂f

∂u′

)

vdx = −∂f

∂u′v|x1 . (2.19)

Note that the above relation holds for every v with v(x0) = 0. If we refine thisclass to both v(x0) = 0 and v(x1) = 0, we see that

x1∫

xo

∂f

∂u−

∂

∂x

(

∂f

∂u′

)

vdx = 0, (2.20)

for all smooth v(x), with v(x0) = 0 = v(x1). From Lemma 2.1.1 then follows

x1∫

xo

∂f

∂u−

∂

∂x

(

∂f

∂u′

)

vdx = 0, ∀v : v(x0) = 0 = v(x1), (2.21)

⇒∂f

∂u−

∂

∂x

(

∂f

∂u′

)

= 0 for x ∈ (x0, x1). (2.22)

Since relation (2.22) holds it follows that equation (2.19) implies for everyv(x0) = 0 (now we drop the condition v(x1) = 0) that

∂f

∂u′(x1, u(x1), u

′(x1)) = 0. (2.23)

This is an additional boundary condition for the differential equation. Itwas not stated in the minimization problem. Since it is only required in the

2.3 A 2-dimensional minimization problem 15

differential equation, the condition is called a natural boundary condition.Hence the minimization problem gives

u(x0) = uo,

∂f

∂u−

∂

∂x(∂f

∂u′) = 0 on (x0, x1),

∂f

∂u′(x1, u(x1), u

′(x1)) = 0.

(2.24)

The boundary condition u(x0) = u0 is called an essential boundary condition:it concerns both the minimization problem and differential equation. The dif-ferential equation (2.22) is also referred to as the Euler-Lagrange equation.

2.3 A 2-dimensional minimization problem

First, we note that Dubois’s Lemma 2.1.1 can also be proved for 2 dimensionaland 3 dimensional situations.

Lemma 2.3.1. Let F (x, y) be continuous over Ω ⊂ R2 with Γ as its boundary,i.e. F ∈ C0(Ω) and suppose that

∫

Ω

F (x, y)v(x, y)dA = 0, ∀v(x, y) ∈ C0(Ω) with v(x, y)|Γ = 0.

ThenF (x, y) = 0 on (x, y) ∈ Ω ∪ Γ.

To treat two dimensional minimization problems, we use the following example:

minu∈U

∫

Ω

1

2‖∇u‖2 − fudA +

∫

Γ2

αuds =: minu∈U

J(u), (2.25)

where U is the set of functions with a continuous derivative on Ω and whereu |Γ1= u1:

U := u smooth : u|Γ1= u1. (2.26)

Further, we assume that the domain of computation Ω is bounded by Γ1 ∪ Γ2,where Γ1 and Γ2 do not overlap. Again we search u ∈ U such that J(u) ≤J(u), ∀u ∈ U . We put J(u + εv) ≥ J(u) ∀v ∈ V := v smooth : v|Γ1= 0(why is v = 0 on Γ1?)

d

dεJ(u + εv)|ε=0 = 0, (2.27)

hence J(u + εv) should be minimal for ε = 0.

2.3 A 2-dimensional minimization problem 16

J(u + εv) =

∫

Ω

1

2∇(u + εv) · ∇(u + εv) − (u + εv)fdA +

∫

Γ2

(u + εv)αds,

=

∫

Ω

1

2∇u · ∇u + ε∇v · ∇u +

1

2ε2∇v · ∇v − fu − εfvdA

+

∫

Γ2

(αu + αεv) ds, (2.28)

d

dεJ(u + εv) =

∫

Ω

∇v · ∇u + ε‖∇v‖2 − fvdA +

∫

Γ2

αvds, (2.29)

d

dεJ(u + εv)|ε=0 =

∫

Ω

∇v · ∇udA −

∫

Ω

fvdA +

∫

Γ2

αvds = 0 for all v |Γ1= 0.

(2.30)

Later we will call equation (2.30) a weak formulation. The Product Rule ofdifferentiation, using

∫

Ω

∇v · ∇udA = −

∫

Ω

v∆udA +

∫

Ω

∇ · (∇uv)dA, (2.31)

(from ∇ · (v∇u) = v∆u + ∇v · ∇u), it follows for equation (2.30):∫

Ω

∇ · [v∇u]dA −

∫

Ω

v∆udA −

∫

Ω

fvdA +

∫

Γ2

αvds = 0. (2.32)

Making use of the two-dimensional Divergence theorem in the first term of theabove equation, gives

∫

Γ1∪Γ2

n · [v∇u]ds +

∫

Γ2

αvds =

∫

Ω

[∆u + f ]vdA ∀v ∈ V. (2.33)

Since u = u1 on Γ1, we have v = 0 on Γ1 and therefore∫

Γ1

v∂u

∂nds = 0, (2.34)

hence∫

Γ2

[∂u

∂n+ α]vds =

∫

Ω

[∆u + f ]vdA ∀v ∈ V. (2.35)

In other words we obtain from the minimization problem:

Find u ∈ U such that∫

Ω

[∆u + f ]vdA =

∫

Γ2

[∂u

∂n+ α]vds ∀v ∈ V. (2.36)

Hence using the DuBois twice:

2.4 Some formalization 17

[1] Take v|Γ2 = 0, then from DuBois’ Lemma it follows

∆u + f = 0;

[2] The previous item implies∫

Γ2

[

∂u

∂n+ α

]

ds = 0,∀v ∈ V,

then, once again DuBois gives the answer

∂u

∂n+ α = 0.

Hence, one obtains

−∆u = f,

u|Γ1 = u1 on Γ1 (Dirichlet),∂u

∂n|Γ2 = −α on Γ2 (Neumann).

(2.37)

The solution of this problem coincides with the minimization problem. TheDirichlet BC is an “essential” BC, the Neumann BC is a new condition ’intro-duced’ by use of DuBois: “Natural BC“. The natural BC did not appear in theminimization problem.

Exercise 2.3.1. Given

J(u) =1

2

∫

Ω

‖∇u‖2dA +

∫

Γ2

1

2gu2 − huds, (2.38)

with u|Γ1= 0 show that J(u) is minimized by the solution of the PDE

∆u = 0 on Ω,∂u

∂n+ gu = h on Γ2,

u|Γ1 = 0.

(2.39)

Hint: use J(u + εv) ≥ J(u) when u is the solution. Set ddε

J(u + εv)|ε=0 = 0.

2.4 Some formalization

Definition 2.4.1. A functional is a map from a real vector-space (of functions)onto R.

Example. Let f(x) be a given function, then

J(u) =

x1∫

x0

(du

dx)2 + fudx, (2.40)


is a functional:

• Input is u = u(x) : a whole curve.

• Ouput ∈ R.

Further f(x) is a known, fixed function of x.

The following theorem answers the question whether the solution of a linear(partial) differential problem is also the solution of a minimization problem.

Theorem 2.4.2. Let L(.) be a linear operator, for instance a minus Laplace op-erator (Lu := −∆u) with homogeneous boundary conditions, then the function,u0, which is the solution of Lu0 = f , minimizes

J(u) =

∫

Ω

(

1

2uLu − uf

)

dA, (2.41)

on U , if the following conditions are satisfied

• L is positive, that is,∫

Ω

uLudA > 0 ∀u ∈ U , for which u 6= 0 for at least

one element of Ω.

• L is self-adjoint (symmetric), that is,∫

Ω

vLudA =∫

Ω

uLvdA ∀u, v ∈ U .

The above remark is proved. We refer to Kreyszig [1989] for a proof formore general cases concerning non-homogeneous boundary conditions.

Proof: We note that

J(u) =

∫

Ω(1

2uLu − uf)dA =

1

2

∫

Ω(u − u0)L(u − u0)dA −

∫

ΩufdA +

1

2

∫

Ω(u0Lu + uLu0 − u0Lu0)dA.

(2.42)Since L is self-adjoint, and Lu0 = f , the above relation can be written as

J(u) =1

2

∫

Ω(u − u0)L(u − u0)dA −

1

2

∫

Ωu0Lu0dA ≥ −

1

2

∫

Ωu0Lu0dA. (2.43)

The last inequality follows since the operator L is positive and equality holdsif u = u0. Hence, u0 minimizes J(u).

In the above theorem, we showed that positivity and self-adjointment are suffi-cient conditions for the existence of a minimization problem. Strictly speaking,if self-adjointness (symmetry) is violated, we did not demonstrate that no min-imization problem exists. However, in practice, positivity and symmetry arealso necessary conditions for the existence of a minimization problem.

Examples:


[1] For −∆u = f a minimization problem can be formulated when u|∂Ω =0 = v|∂Ω.

[2] For ∆u − q · ∇u = f no minimization problem can be formulated foru|∂Ω = 0.

Reasons:

[1]

−

∫

Ω

v∆udA = −

∫

Ω

∇ · [v∇u]dA +

∫

Ω

∇v · ∇udA, (2.44)

= −

∫

∂Ω

n · v∇udS +

∫

Ω

∇v · ∇udA, (2.45)

= +

∫

Ω

∇v · ∇udA (since u|∂Ω = 0 = v|∂Ω), (2.46)

−

∫

Ω

u∆vdA = −

∫

Ω

∇ · [u∇v]dA +

∫

Ω

∇u · ∇vdA, (2.47)

= +

∫

Ω

∇u∇vdA (since u|∂Ω = 0 = v|∂Ω). (2.48)

Hence −∆(·) is self-adjoint, i.e.

∫

Ω

v∆udA =

∫

Ω

u∆vdA. (2.49)

And

−

∫

Ω

u∆udA =

∫

Ω

‖ ∇u ‖2 dA ≥ 0, (2.50)

hence −∆(·) is positive.

[2] Let q be a given constant vector and u|∂Ω = 0 = v|∂Ω, then

∫

Ω

vq · ∇udA =

∫

∂Ω

uvq · ndS −

∫

Ω

uq · ∇vdA = −

∫

Ω

uq · ∇vdA. (2.51)

Hence,∫

Ω

vq · ∇udA 6=

∫

Ω

uq · ∇vdA, (2.52)

(not self-adjoint).

3

Variational Formulation andDifferential equations

In the previous chapter we saw the relation between minimization problemsand (partial) differential equations. It was demonstrated that if the differentialoperator is positive and self-adjoint, then, such a minimization problem exist.Ritz’ finite element method is based on the numerical solution of a minimizationproblem. To solve problems with differential operators that do not satisfy theserequirements, the so-called weak form is introduced. The differential equationis written as a weak form and then a numerical solution to this weak form isdetermined. This method is more generally applicable and it is the backboneof Galerkin’s finite element method.

3.1 Weak forms

Consider the following minimization problem on domain Ω with boundaries∂Ω = Γ1 ∪ Γ2:

Find u smooth, such that u|Γ1 = g, and

J(u) ≤ J(u) for all smooth u with u|Γ1 = g,

where J(u) := 12

∫

Ω

‖∇u‖2dA.(3.1)

Using u = u + εv for all smooth v with v|Γ1 = 0, it follows that the solution ofequation (3.1) coincides with:

Find u smooth, such that u|Γ1 = g, and∫

Ω

∇u.∇vdA = 0 for all smooth v with v|Γ1 = 0. (3.2)

It can be demonstrated that the solution of the above exists and that it isunique. We suppose that the boundary of Ω is given by Γ1 ∪Γ2. The problems

3.1 Weak forms 21

(3.1) and (3.2) have the same solution. The Product Rule for differentiationapplied to

∫

Ω

∇u · ∇vdA gives:

∫

Ω

∇ · [v∇u]dA −

∫

Ω

v∆udA = 0 ∀v with v|Γ1 = 0, (3.3)

or∫

Γ1

v∂u

∂nds +

∫

Γ2

v∂u

∂nds =

∫

Ω

v∆udA ∀v|Γ1 = 0 (with v smooth), (3.4)

with v|Γ1 = 0, follows

∫

Γ2

v∂u

∂ndS =

∫

Ω

v∆udA ∀v|Γ1 = 0 with v smooth. (3.5)

Suppose that v|Γ2 = 0 besides v|Γ1 = 0, then Du Bois’s Lemma gives ∆u = 0on Ω. When we release v|Γ2 = 0 and use ∆u = 0 on Ω, we obtain the following

natural boundary condition∂u

∂n|Γ2 = 0. Hence smooth solutions of (3.1) and

(3.2) satisfy:

−∆u = 0,

u|Γ1 = g,∂u∂n

|Γ2 = 0.

(3.6)

When ∆u exists within Ω, then the solutions of equations (3.1), (3.2) and (3.6)are the equal. (3.6) contains a PDE, (3.1) is its corresponding minimizationproblem and (3.2) is called a ’variational formulation’ or a ’weak form’ of PDE(3.6).

So far we went from a weak form to a problem with a PDE. In practice, oneoften goes the other way around. Since Finite Element Methods are based oneither the solution of a minimization problem (such as (3.1)) or a weak form(as in (3.2)), we would like to go from a PDE to a weak from. Further, thecondition v|Γ1 = 0 because of u|Γ1 = g and hence prescribed, originates fromthe use of a minimization problem.

Solving of (3.6) by a numerical solution of the representation of (3.2) isreferred to as Galerkin’s method. Whereas, aquiring the numerical solutionof a representation of (3.1) is called Ritz’ method. Galerkin’s method is mostgeneral: it can always be applied. It doesn’t matter whether differential op-erators are self-adjoint or positive. Therefore, this method will be treated inmore detail. The study of minimization problems was needed to motivate thecondition v = 0 on locations where u is prescribed (by an essential condition).

A mayor advantage of the weak form (3.2) is the fact it is easier to proveexistence and uniqueness of (3.2) than for one satisfying (3.6) with boundaryconditions. It is clear that a solution of (3.6) always is always a solution of(3.2). A solution of the PDE (3.6) always needs the second order derivatives to

3.1 Weak forms 22

exist, whereas in the solution of (3.2) only the integrals have to exist. For thesolutions of the weak form, it may be possible that the second order derivativesdo not exist at all. For that reason, the term weak form or weak solution is usedfor the problem and its solution respectively.

The function v is commonly referred to as a ’test function’. Let’s go from(3.6) to (3.2). Given ∆u = 0 ⇔ v∆u = 0 for all v|Γ1 = 0 (reason is thatu|Γ1 = g is prescribed!) then

∫

Ω

v∆udA = 0, (3.7)

∫

Ω

∇ · [v∇u]dA −

∫

Ω

∇u · ∇vdA = 0 ∀v|Γ1 = 0. (3.8)

The Product Rule for differentiation was used here. Using the DivergenceTheorem, this gives (since v|Γ1 = 0)

∫

Γ2

v∂u

∂nds −

∫

Ω

∇u · ∇vdA = 0 ∀ v|Γ1 = 0. (3.9)

Since in (3.6) it is required that∂u

∂n|Γ2 = 0 , we obtain

∫

Γ2

v∂u

∂nds = 0, (3.10)

and hence,∫

Ω

∇u · ∇vdA = 0 ∀ v|Γ1 = 0 smooth. (3.11)

Hence (3.6) is equivalent to (3.2), if we are not bothered by the smoothnessconsiderations:

Find u smooth, such that u|Γ1 = g, and∫

Ω

∇u · ∇vdA = 0 for all smooth v with v|Γ1 = 0. (3.12)

Here (3.2) is also sometimes referred to as the ’Finite Element’ formulation of(3.6). The same principle may be applied to

∂c

∂t= ∆c + f,

c|Γ1 = g,∂c

∂n|Γ2 = h,

c(x, y, 0) = 0, t = 0 (x, y) Ω.

(3.13)

In the above problem the boundary of the domain of computation Ω is given byΓ1∪Γ2. The question is now to find a Finite Element formulation for (3.13). We

3.1 Weak forms 23

multiply the PDE with a testfunction v, that satisfies v|Γ1 = 0, since c|Γ1 = gis prescribed, to obtain, after integration over Ω,

∫

Ω

∂c

∂tvdA =

∫

Ω

v∆cdA +

∫

Ω

fvdA ∀v |Γ1= 0, (3.14)

(v smooth). By use of the productrule for differentiation, we obtain

∫

Ω

∂c

∂tvdA =

∫

Ω

∇ · [v∇c]dA −

∫

Ω

∇c · ∇vdA +

∫

Ω

fvdA, ∀v |Γ1= 0. (3.15)

The Divergence Theorem implies:

∫

Ω

∂c

∂tvdA =

∫

Γ1

∂c

∂nvds+

∫

Γ2

∂c

∂nvds−

∫

Ω

∇c·∇vdA+

∫

Ω

fvdA, ∀v |Γ1= 0. (3.16)

Since∂c

∂n= h on Γ2 and v |Γ1= 0, we obtain: Find c with c |t=0= 0, c |Γ1= g

such that∫

Ω

∂c

∂tvdA =

∫

Γ2

hvds −

∫

Ω

∇c · ∇vdA +

∫

Ω

fvdA, ∀v |Γ1= 0. (3.17)

Equation (3.17) is the variational form or Finite Element Form of (3.13). Notethat the Neumann BC is changed into a line-integral over Γ2. Of course, it iseasy to show that (3.13) can be derived, once only (3.17) is given:

∫

Ω

[

∂c

∂t− f

]

vdA =

∫

Γ2

hvds −

∫

Ω

∇ · [v∇c]dA +

∫

Ω

v∆cdA, ∀v |Γ1= 0. (3.18)

Using v |Γ1= 0, this gives

∫

Ω

[

∂c

∂t− ∆c − f

]

vdA =

∫

Γ2

[

h −∂c

∂n

]

vds ∀v |Γ1= 0. (3.19)

If we set v |Γ2= 0 besides v |Γ1= 0, we obtain from DuBois

∫

Ω

[

∂c

∂t− ∆c − f

]

vdA = 0 ⇒∂c

∂t− ∆c − f = 0 on Ω. (3.20)

This implies after releasing v |Γ2= 0 that h −∂c

∂n= 0 on Γ2 (again from

DuBois). We see that (3.17) corresponds with (3.13), since we require c |Γ1= gand c |t=0= 0 for both (3.13) and (3.17). Note that for the derivation of theweak form, we always multiply the PDE with a test-function v, which mustsatisfy v = 0 on a boundary with a Dirichlet condition (i.e. an essential condi-tion). Subsequently we integrate over the domain of computation.

3.2 Which weak formulation? 24

Exercise 3.1.1. Suppose that we have been given the following problem:

∂c

∂t= ∆c on Ω,

c |Γ1= f on Γ1,

c |Γ2 +∂c

∂n|Γ2 = g on Γ2,

c(x, y, 0) = 0 for t = 0 on Ω.

(3.21)

Show that a weak form of the above problem is given by:Find c smooth, subject to c |Γ1= f and c |t=0= 0, such that

∫

Ω

∂c

∂tvdA = −

∫

Ω

∇c · ∇vdA +

∫

Γ2

(g − c) vds ∀v |Γ1= 0. (3.22)

v smooth.The above weak form (3.22) is used to solve (3.21) by the use of Finite Ele-ments. Note that the Robin-condition is a natural boundary condition, which iscontained in the weak form in the second term of the right-hand of the (3.22).

3.2 Which weak formulation?

When we considered

∆u = f on Ω,

u |Γ= 0,(3.23)

then we saw that a weak form is given by (2.30)

Find u |Γ= 0 such that,

−

∫

Ω

∇u · ∇vdA =

∫

Ω

fvdA for all v |Γ= 0. (3.24)

The above problem is a weak form, but the following problem is also a weakform:

Find u |Γ= 0 such that,∫

Ω

v∆udA =

∫

Ω

fvdA for all v |Γ= 0, (3.25)

or even

Find u |Γ= 0 such that,

−

∫

Ω

u∆vdA =

∫

Ω

fvdA for all v |Γ= 0. (3.26)

Forms (3.24),(3.25) and (3.26) are all possible weak forms of (3.23). However, inthe Finite Element calculations, (3.25) and (3.26) are not common. This is dueto the reduction of order of the derivatives in the first form (3.24). Here only thefirst derivatives are used and this will give an advantage for the implementation

3.3 Mathematical considerations:existence and uniqueness 25

of the FEM, which we will see later. A more important advantage is that for aminimized order of derivatives in the weak form, the class of allowable solutionsis largest, in the sense that solutions that are less smooth are allowable. Asa rule of thumb, now, we mention that when we derive a weak form, then weshould try to minimize the highest order of the derivatives that occur in theintegrals.Example:

u′′′′ = f f on x ∈ (0, 1),

u(0) = 0,

u′(0) = 0,

u(1) = 0,

u′(1) = 0.

(3.27)

We derive a weak form with the lowest order for the derivatives.

1∫

0

u′′′′vdx =

1∫

0

fvdx for all v(0) = 0 = v′(0) = v(1) = v′(1), (3.28)

partial integration gives

[

u′′′v]1

0−

1∫

0

u′′′v′dx =

1∫

0

fvdx for all v(0) = 0 = v′(0) = v(1) = v′(1), (3.29)

with the condition v(0) = 0 = v(1) follows

−

1∫

0

u′′′v′dx =

1∫

0

fvdx for all v(0) = 0 = v′(0) = v(1) = v′(1). (3.30)

Partial integration, again, gives (with v′(1) = 0 = v′(0))

1∫

0

u′′v′′dx =

1∫

0

fvdx for all v(0) = 0 = v′(0) = v(1) = v′(1). (3.31)

Now we stop, because, another partial integration would increase the maximumorder of the derivatives again to obtain

1∫

0

u′v′′′dx =

1∫

0

fvdx for all v(0) = 0 = v′(0) = v(1) = v′(1). (3.32)

We do not use (3.32) but (3.31) as the weak form for the Finite Element Method.

3.3 Mathematical considerations:existence and uniqueness

This section is intended for the interested reader and it is not necessary forthe understanding of the finite element method. Consider the following Poisson


problem

−∆u = f(x), for x ∈ Ω,

u = g(x), for x ∈ ∂Ω.(3.33)

Here we assume that f(x) and g(x) are given continuous functions. Let Ω =Ω ∪ ∂Ω be the closure of Ω, then the following assertion can be demonstrated

Theorem 3.3.1. If f(x) and g(x) are continuous and if the boundary curveis (piecewise) smooth, then problem (3.33) has one and only one solution suchthat u ∈ C2(Ω) ∩ C1(Ω) (that is the solution has continuous partial derivativesup to at least the second order over the open domain Ω and at least continuousfirst order partial derivatives on the boundary).

We will not prove this result for the existence and uniqueness of a classicalsolution to problem (3.33). The proof of the above theorem is far from trivial,the interested reader is referred to the monograph of Evans for instance. Thefact that the second order partial derivatives need to be continuous is a ratherstrong requirement.

The finite element representation of the above problem is given by

Find u ∈ H1(Ω), subject to u = g on ∂Ω, such that

∫

Ω ∇u · ∇φdΩ =∫

Ω φfdΩ, for all φ ∈ H1(Ω).

(3.34)

In the above problem (4.86), the notation H1(Ω) has been used, this concernsthe set of functions for which the integral over Ω of the square of the functionand its gradient is finite. Informally speaking, this is

u ∈ H1(Ω) ⇐⇒

∫

Ωu2dΩ < ∞ and

∫

Ω||∇u||2dΩ < ∞. (3.35)

This set of functions represents a Hilbert space and is commonly referred toas a Sobolev space. Using the fact that each function that is in H1(Ω) is alsocontinuous on Ω, that is H1(Ω) ⊂ C0(Ω), the following claim can be proved

Theorem 3.3.2. The problem (4.86) has one and only one solution u, suchthat u ∈ H1(Ω).

The proof of the above theorem resides on the Lax-Milgram Theorem (seefor instance the book by Kreyszig [1989]):

Theorem 3.3.3. Let V be a Hilbert space and a(·, ·) a bilinear form on V,which is

[1] bounded: |a(u, v)| ≤ C‖u‖‖v‖ and

[2] coercive: a(u, u) ≥ c‖u‖2.

Then, for any linear bounded functional f ∈ V ′, there is a unique solutionu ∈ V to the equation

a(u, v) = f(v), for all v ∈ V. (3.36)


The proof takes into account the fact that the linear operator is positive(more exactly speaking coercive, which is

∫

Ω ||∇u||2dΩ ≥ α∫

Ω u2dΩ for someα > 0) and continuous. Further, the right hand side represents a boundedlinear functional. These issues constitute the hypotheses under which the Lax-Milgram theorem holds and hence have to be demonstrated. In this text the(straighforward) proof is omitted and a full proof of the above theorem can befound in van Kan [2006] for instance.

The most important lesson that we learn here, is that the solution to theweak form exists and that it is uniquely defined. Further, the weak form allowsa larger class of functions as solutions than the PDE does.

4

Galerkin’s Finite Element method

In this chapter we treat the Finite Element Method, which was proposed byGalerkin. The method is based on the weak form of the PDE. This Galerkinmethod is more general than the Ritz’ method, which is based on the solutionof a minimization problem and, hence, is only suitable whenever the differentialoperator is positive and self-adjoint.

4.1 The principle of Galerkin’s method

Given the following weak form:

Find u |Γ= 0 such that,∫

Ω

∇u · ∇vdA = 0 for all v |Γ= 0. (4.1)

Here Ω is a general simply connected domain in R1 or R2 or R3. A crucialprinciple for the FEM is that we write u as a sum of “basis-functions” ϕi(x, y),which satisfy

ϕi(x, y) |Γ= 0, (4.2)

i.e. hence

u(x, y) =

∞∑

j=1

cjϕj(x, y). (4.3)

Since, we cannot do calculations with an infinite number of terms, we truncatethis series such that we only take the first n terms into account, then

u(x, y) =

n∑

j=1

cjϕj(x, y), (4.4)

4.1 The principle of Galerkin’s method 29

where u denotes the approximation of the solution of (4.1). As an examplefor ϕi one might take powers, sines, cosines (viz. Fourier) and so on. We willassume here that

u(x, y) =n∑

j=1

cjϕj(x, y) → u(x, y) as n → ∞, (4.5)

note that u(x, y) represents the approximated solution of (4.1) and u(x, y) theexact solution of (4.1) respectively. There is a lot of mathematical theory neededto prove that u → u as n → ∞ for a specific set of basis-functions ϕi(x, y). TheFinite Element representation of weak form (4.1) is:

Find the set of constants c1, . . . , cn such that,n∑

j=1

∫

Ω

cj∇ϕj(x, y) · ∇ϕi(x, y)dA = 0 for all i ∈ 1, . . . , n. (4.6)

Note that we assume here that all functions v |Γ= 0 are represented by (linearcombinations of) the set ϕi(x, y), i ∈ 1, . . . , n. We will use this assumptionand skip the mathematical proof (see Strang and Fix [1973], Cuvelier et al.[1986] and Braess [1996] for instance for a proof). Further, in (4.6), we willmake a choice for the functions ϕi(x, y) and hence they are known in theFinite Element calculations. It turns out that the choice of the basis-functionsϕi(x, y) influences the accuracy and speed of computations. The accuracy isa difficult subject, which we will treat without detail. The speed of computationis easier to deal with. Note that u |Γ= 0 due to ϕi(x, y) |Γ= 0, i ∈ 1, . . . , n.(4.6) implies a set of linear equations of ci:

c1R

Ω

∇ϕ1 · ∇ϕ1dA + c2R

Ω

∇ϕ2 · ∇ϕ1dA + c3R

Ω

∇ϕ3 · ∇ϕ1dA + · · · + cn

R

Ω

∇ϕn · ∇ϕ1dA = 0,

c1R

Ω

∇ϕ1 · ∇ϕ2dA + c2R

Ω

∇ϕ2 · ∇ϕ2dA + c3R

Ω

∇ϕ3 · ∇ϕ2dA + · · · + cn

R

Ω

∇ϕn · ∇ϕ2dA = 0,

..

....

c1R

Ω

∇ϕ1 · ∇ϕndA + c2R

Ω

∇ϕ2 · ∇ϕndA + c3R

Ω

∇ϕ3 · ∇ϕndA + · · · + cn

R

Ω

∇ϕn · ∇ϕndA = 0,

(4.7)

The ’discretization’ matrix here is refered to as the stiffness matrix, its elementsare

Aij =

∫

Ω

∇ϕi · ∇ϕjdA (4.8)

Exercise 4.1.1. Show that A is symmetric, i.e. aij = aji.

For a fast solution of the system of linear equations, one would like A to beas sparse as possible (i.e. A should contain as many zeros as possible). Whenϕi are orthogonal over the ∇, then

∫

Ω

∇ϕi · ∇ϕjdA = 0 when i 6= j. (4.9)

4.1 The principle of Galerkin’s method 30

This would be ideal: the cj then follows very easily:

cj =0

∫

Ω

∇ϕj∇ϕj= 0, (4.10)

(Of course when A is not singular the solution of the above system of equationis given by ci = 0, we here only illustrate the working of the FEM)

In practice it is not always possible to choose a set of orthogonal basis-functions, but we try to choose a set that is almost orthogonal. This meansthat A consists of zeroes mainly (i.e. A is a sparse matrix). We will choosebasis-functions ϕi(x, y) that are piecewise linear. Suppose that the domainof computation is divided into a set of gridnodes, see below, with numbers forthe unknowns (figure 4.1).

6

8

10

11

12

12

14

17

15

19

20 25

24

23

22

2116

18

7

9

5

4

3

2

1

Figure 4.1: An example of a domain divided into a Finite Element mesh

Then, we will choose ϕi(x, y) to be piecewise (bi-)linear, such that

ϕi(xj , yj) =

1, for (xj , yj) = (xi, yi),

0, for (xj , yj) 6= (xi, yi).(4.11)

The reason for this choice will be motivated by the use of a one-dimensionalexample. It is clear that for this case the integrals of,

∫

Ω

∇ϕi · ∇ϕjdA, (4.12)

only do not vanish when i and j are equal or when i and j are neighboringgridpoints, due to piecewise linearity of the basis-functions ϕi. This impliesthat the basis-functions have a compact support and hence, the stiffness-matrixwill be sparse. First we motivate the choice of piecewise linear basis-functionsby the use of a 1-Dimensional example.

4.2 Motivation of piecewise linear basis-functions 31

xi+2xi+1xixi−1xi−2

u(x)

x

Figure 4.2: The function u(x) at gridnodes xi

1

l(x)

xixi−1

li(x)

li−1(x)

x

1

l(x)li(x)

li−1(x)

li(x)

xi−1 xi xi+1

(b)(a)

Figure 4.3: The piecewise linear functions li(x)

4.2 Motivation of piecewise linear basis-functions

Given any function u = u(x) and a division of gridnodes, see figure 4.2. For[xi−1, xi], we approximate u(x) by the use of linear interpolation:

u(x) = u(xi−1) +u(xi) − u(xi−1)

xi − xi−1(x − xi−1) for x ∈ [xi−1, xi], (4.13)

or with ui = u(xi), we obtain (figure 4.3)

u(x) = ui−1 +ui − ui−1

xi − xi−1(x − xi−1),

= ui−1

1 +xi−1 − x

xi − xi−1

+ ui

x − xi−1

xi − xi−1

,

= ui−1xi − x

xi − xi−1+ ui

x − xi−1

xi − xi−1=: ui−1li−1(x) + uili(x). (4.14)

Henceu(x) = ui−1li−1(x) + uili(x) for x ∈ [xi−1, xi]. (4.15)

4.3 Evaluation of a one-dimensional example 32

1

xi−1 xi xi+1

x

Ψi(x)

Ψ

Figure 4.4: The piecewise linear function φi(x)

We do the same for x ∈ [xi, xi+1] (see Figure 4.3), to obtain:

u(x) = uili(x) + ui+1li+1(x) on x ∈ [xi, xi+1]. (4.16)

Therewith, we write for the approximation u(x) of u(x):

u(x) =

n∑

j=1

ujφj(x), (4.17)

where

φi(x) =

x − xi−1

xi − xi−1, for x ∈ [xi−1, xi],

x − xi+1

xi − xi+1, for x ∈ [xi, xi+1],

0, for x /∈ [xi−1, xi+1].

(4.18)

Hence φi(x) is piecewise linear, see Figure 4.4, where

φi(xj) =

0, i 6= j,

1, i = j.(4.19)

In the Finite Element Method we put in the functions φi(x) and we determineui. Therewith, we obtain the solution.

4.3 Evaluation of a one-dimensional example

Now we treat procedures to approximate the integrals. For this set of basis-functions we solve:Find u(x) such that u(0) = 0 and

−

1∫

0

u′(x)v′(x)dx =

∫ 1

0f(x)v(x)dx, for all v(0) = 0. (4.20)

4.3 Evaluation of a one-dimensional example 33

Exercise 4.3.1. Find the differential equation for u with all boundary condi-tions that corresponds to (4.20) with u(0) = 0.

The approximation u(x) =

n∑

j=1

ujφj(x) (where u → u as n → ∞ is assumed)

leads to the following Finite Element formation: Find coefficient ui, in u(x) =

n∑

j=1

ujφj(x),

with

φi(xj) =

0, i 6= j,

1, i = j,(4.21)

such that

−

1∫

0

n∑

j=1

ujφ′j(x)φ′

i(x)dx =

∫ 1

0f(x)φi(x)dx ∀i ∈ 1, . . . , n. (4.22)

The computation of

1∫

0

φ′i(x)φ′

j(x)dx is simple, since φi are piecewise linear

functions

1∫

0

φ′i−1(x)φ′

i(x)dx =

xi∫

xi−1

1

xi−1 − xi

1

xi − xi−1dx = −

1

xi − xi−1, (4.23)

1∫

0

φ′i(x)φ′

i(x)dx =1

xi − xi−1+

1

xi+1 − xi, (4.24)

1∫

0

φ′i(x)φ′

i+2(x)dx = 0, (4.25)

(why?). Of course, given any function f(x) whose product with φi(x) is in-

tegrable, the integral

1∫

0

f(x)φi(x)dx may be evaluated. For some cases it is

even possible to find an anti-derivative for φi(x)f(x). However, for many casesthis isn’t possible. We then evaluate the integral numerically. Standard FE-

4.4 Ritz’ method of Finite Elements by a simple example 34

packages use the Newton-Cotes formulas, which read as

1∫

0

f(x)φi(x)dx = f(xi)

1∫

0

φi(x)dx

= f(xi)

xi+1∫

xi−1

φi(x)dx

= f(xi)1

2(xi+1 − xi−1) for i ∈ 1, 2, . . . , n − 1.(4.26)

For i = n we use1∫

0

f(x)φn(x)dx =

xn∫

xn−1

f(xn)φn(x)dx = f(xn)1

2(xn − xn−1), (4.27)

or better1

2f

(

xn + xn−1

2

)

(xn − xn−1). (4.28)

Evaluation of all integrals in (4.22) gives a system of linear equations in ui.We also know that u′(1) = 0 results from (4.20) as a natural (BC). This condi-tion should also follow from (4.22):

1∫

0

φ′n−1φ

′ndx = −

1

xn − xn−1, (4.29)

1∫

0

φ′nφ′

ndx =1

xn − xn−1, (4.30)

⇒ −1

xn − xn−1un−1 +

1

xn − xn−1un = f(xn)

1

2(xn − xn−1), (4.31)

where xn =xn + xn−1

2can be chosen (or just xn = xn). Further f(xn) = u′′(xn)

(see exercise 4.3.1). Implies

un − un−1

xn − xn−1=

1

2f(xn)(xn − xn−1)

=1

2u′′(xn)(xn − xn−1). (4.32)

This implies thatun − un−1

xn − xn−1→ 0 as n → ∞ (xn−xn−1 → 0). Hence the natural

(BC) is recovered.

4.4 Ritz’ method of Finite Elements by a simple example

Suppose that we have the following minimization problem:

minu∈U

J(u),with J(u) =1

2

1∫

0

(

du

dx

)2

dx, (4.33)

4.5 The treatment of a non-homogeneous Dirichlet BC 35

where U := u smooth: u(0) = 0. Then this problem corresponds to thesolution of

d2u

dx2= 0 on (0, 1),

u(0) = 0,du

dx(1) = 0.

(4.34)

Note that the solution is given by u = 0. Now we use Ritz’ FEM to solve theminimazation problem. Let the approximate solution be given by (figure 4.4)

un(x) =

n∑

j=1

ujφj(x), (4.35)

where we assume that un(x) → u(x) on x ∈ [0, 1] as n → ∞. Then we look forconstants u1, u2, . . . , un such that J(u) is minimal. In other words:

∂

∂uiJ(u) = 0 ∀i ∈ 1, 2, 3, . . . , n. (4.36)

Substitution of

n∑

j=1

ujφj(x) = un(x) into J(u) gives:

J(u) =1

2

1∫

0

n∑

i=1

uiφ′i(x) ·

n∑

j=1

ujφ′j(x)dx. (4.37)

Hence

∂

∂uiJ(u) = 0 ⇒

n∑

j=1

1∫

0

ujφ′j(x)φ′

i(x)dx = 0 ∀i ∈ 1, . . . , n. (4.38)

This gives exactly the same equations as in (4.22). Note the similarity with theresult obtained by the use of Galerkin’s method. Since Galerkin’s method isapplicable for more general cases we do not treat this method further.

4.5 The treatment of a non-homogeneous Dirichlet BC

Suppose that we have to solve the following variational problem:

Find u(x), subject to u(0) = u0, such that

−

1∫

0

u′(x)v′(x)dx =

1∫

0

f(x)v(x)dx ∀v(0) = 0. (4.39)

Now the essential condition is non-zero (when u0 6= 0), where u0 is given.The treatment is similar to the case where u0 = 0, but now we set

un(x) =n∑

j=0

ujφj(x) = u0φ0(x) +n∑

j=1

ujφj(x). (4.40)

4.5 The treatment of a non-homogeneous Dirichlet BC 36

1

xi−1 xi xi+1

x

Ψ

Ψi(x)

1 Ψ0(x)

x10

Figure 4.5: The piecewise linear functions φi(x) and φ0(x)

Note that u0 is from the Dirichlet BC. Again, we use piecewise linear basis-functions φi(x), where

φi(xj) =

1 i = j,

0, i 6= j,and φi(x) is piecewise linear. (4.41)

For φ0(x) we have: φ0(x0) = φ0(0) = 1. For the functions v(x) we set φi(x)where i ∈ 1, . . . , n (note that φi(0) = 0 since v(0) = 0). Then, we obtain thefollowing problem:

Find u1, u2, . . . , un such that

−

1∫

0

n∑

j=0

ujφ′j(x)φ′

i(x)dx =

1∫

0

f(x)φi(x)dx, (4.42)

for all φi(x)ni=1, i.e. ∀i ∈ 1, . . . , n.

For the functions φi(x) we use the sketches from figure 4.5. For φ1(x), wherej = 1, follows (where u0 is prescribed):

−

1∫

0

u0φ′0φ

′1dx −

1∫

0

u1φ′1φ

′1dx −

1∫

0

u2φ′2φ

′1dx =

1∫

0

f(x)φ1dx, (4.43)

−1

x1 − x0−

1

x2 − x1

u1 +1

x2 − x1u2

= −u0

x1 − x − 0+ f(x1)

1

2(x2 − x0).(4.44)

The integrals can also be evaluated for the other values of j. The same equationsfollow as in the example where u0 = 0. Again, we will have that u′

n(1) → 0 asn → ∞. For more dimensional problems, the same occurs: Suppose that wesearch u, subject to u |Γ2= g, such that

∫

Ω

∇u · ∇vdA = 0 ∀v |Γ= 0. (4.45)

4.6 A time-dependent example 37

Let n be the number of gridnodes that are inside Ω (figure 4.5) or on boundaryΓ2, (but not on Γ1), and let n + 1, . . . , n + m be the gridnodes on boundaryΓ1, then we set

u(x, y) ≈n∑

j=1

ujφj(x, y) +

n+m∑

j=n+1

g(xj , yj)φj(x, y). (4.46)

Here we take

φj(xi, yi) =

1, for j = i,

0, for j 6= i,(4.47)

where φj(x, y) are taken to be piecewise linear. Expression (4.46) is substitutedinto the weak form to obtain a system of linear equations.

4.6 A time-dependent example

Consider the following problem:

∂c

∂t=

∂2c

∂x2on x ∈ (0, 1), t > 0,

c(0, t) = 1 t > 0,∂c

∂x(1, t) = 0 t > 0,

c(x, 0) = 0 on x ∈ (0, 1).

(4.48)

First we search a weak form for (4.48): Search c, c(0, t) = 1, c(x, 0) = 0,

1∫

0

∂c

∂tvdx =

1∫

0

∂2c

∂x2vdx ∀v(0, t) = 0. (4.49)

To reduce the order of the derivative, we integrate by parts:

1∫

0

∂c

∂tvdx =

[

∂c

∂xv

]1

0

−

1∫

0

∂c

∂x

∂v

∂xdx = −

1∫

0

∂c

∂x

∂v

∂xdx, (4.50)

since v(0, t) = 0 and∂c

∂x(1, t) = 0. Then, we obtain the following weak form

Find c, subject to c(0, t) = 1, c(x, 0) = 0, such that1∫

0

∂c

∂tvdx = −

1∫

0

∂c

∂x

∂v

∂xdx ∀v(0) = 0.

(4.51)

We solve (4.51) by use of the Galerkin FEM. Again, we use piecewise linearbasis-functions φi(x)n

i=0 as before on the n gridnodes, with

φi(xj) =

1, j = i,

0, j 6= i.(4.52)

4.7 The principle of element matrices and vectors 38

Then, we approximate c(x, t) by

cn(x, t) =

n∑

j=0

cj(t)φj(x) = φ0(x) +

n∑

j=1

φj(x)cj(t). (4.53)

Note that cj should be functions of x and t since c(x, t) is a function of t andφj(x) is a function of x only. Substitution into (4.51) gives

Find ci(t)ni=1 such that

1∫

0

n∑

j=1

c′j(t)φj(x)φi(x)dx

= −

1∫

0

φ′0(x) +

n∑

j=1

φ′j(x)cj(t)

φ′i(x)dx ∀i ∈ 1, . . . , n.

(4.54)

The above problem represents a system of linear ordinary differential equations.

Note that c′j(t) ≈cj(t + ∆t) − cj(t)

∆t. We will deal with the solution of the time-

dependent problem in the next chapter.

4.7 The principle of element matrices and vectors

We briefly treat the concept of element matrices, which is not a mathematicalfeature but only a convenient programming trick. This trick is used in most ofthe implementations of Finite Elements software.

For the treatment we treat a simple one-dimensional example:

−u′′ = f,

u(0) = u0,

u′(1) = 0.

(4.55)

A weak form is then obtained by

−u′′v = fv ⇔ −

1∫

0

u′′vdx =

1∫

0

fvdx

⇔[

−u′v]1

0+

1∫

0

u′v′dx =

1∫

0

fvdx

⇔

1∫

0

u′v′dx =

1∫

0

fvdx. (4.56)

We use Galerkin’s method on n gridpoints for the unknowns:

u(x) =n∑

j=1

ujφj(x) + u0φ0(x). (4.57)


Substitution into the weak form, with

φj(xi) =

1, j = i,

0, j 6= i,(4.58)

piecewise linear, gives

1∫

0

n∑

j=0

ujφ′j(x)φ′

i(x)dx =

1∫

0

f(x)φi(x)dx, i ∈ 1, . . . , n, (4.59)

or

n∑

j=0

uj

1∫

0

φ′j(x)φ′

i(x)dx =

1∫

0

f(x)φi(x)dx − u0

1∫

0

φ′0(x)φ′

i(x)dx i ∈ 1, . . . , n.

(4.60)In other words:

u1

1∫

0

φ′1φ

′1dx + u2

1∫

0

φ′2φ

′1dx + · · · + un

1∫

0

φ′nφ′

1dx =

1∫

0

fφ1 − u0

1∫

0

φ′0φ

′1dx,

u1

1∫

0

φ′1φ

′2dx + u2

1∫

0

φ′2φ

′2dx + · · · + un

1∫

0

φ′nφ′

2dx =

1∫

0

fφ2 − u0

1∫

0

φ′0φ

′2dx,

...

For simplicity we take n=3; then,

u1

1∫

0

φ′1φ

′1dx + u2

1∫

0

φ′2φ

′1dx + u3

1∫

0

φ′3φ

′1dx =

1∫

0

fφ1 − u0

1∫

0

φ′0φ

′1dx,

u1

1∫

0

φ′1φ

′2dx + u2

1∫

0

φ′2φ

′2dx + u3

1∫

0

φ′3φ

′2dx =

1∫

0

fφ2 − u0

1∫

0

φ′0φ

′2dx,

u1

1∫

0

φ′1φ

′3dx + u2

1∫

0

φ′2φ

′3dx + u3

1∫

0

φ′3φ

′3dx =

1∫

0

fφ3 − u0

1∫

0

φ′0φ

′3dx.

(4.61)

We take piecewise linear basis-functions. Note that we have

1∫

0

φ′1φ

′3dx = 0

(the basis-functions are called nearly orthogonal) for instance and that onlycontributions from neighbouring elements are non-zero. Next, we will introducethe concept element as an interval between adjacent meshpoint: ei := [xi−1, xi].


Hence, using the fact that the basis-functions are nearly orthogonal, we writethe above equations as follows:

u1

∫

e1∪e2

φ′1φ

′1dx + u2

∫

e2

φ′1φ

′2dx =

∫

e1∪e2

fφ1dx − u0

∫

e1

φ′0φ

′1dx,

u1

∫

e2

φ′2φ

′1dx + u2

∫

e2∪e3

φ′2φ

′2dx + u3

∫

e3

φ′2φ

′3dx =

∫

e2∪e3

fφ2dx,

u2

∫

e3

φ′2φ

′3dx + u3

∫

e3

φ′3φ

′3dx =

∫

e3

fφ3dx. (4.62)

Note further that

∫

ei∪ei+1

φ′iφ

′idx =

∫

ei

φ′iφ

′idx +

∫

ei+1

φ′iφ

′idx

∫

ei∪ei+1

fφidx =

∫

ei

fφidx +

∫

ei+1

fφidx(4.63)

Now we show a computer procedure to generate the stiff-ness matrix (discretiza-tion matrix) by use of element-matrices: start with,

A0 =

0 0 0

0 0 0

0 0 0

, (4.64)

we finish with

A =

∫

e1

φ′1φ

′1 +

∫

e2

φ′1φ

′1

∫

e2

φ′1φ

′2 0

∫

e2

φ′2φ

′1

∫

e2

φ′2φ

′2 +

∫

e3

φ′2φ

′2

∫

e3

φ′2φ

′3

0∫

e3

φ′2φ

′3

∫

e3

φ′3φ

′3

. (4.65)

Now we introduce the concept of element-matrices:

sei=

∫

ei

φ′i−1φ

′i−1dx

∫

ei

φ′i−1φ

′idx

∫

ei

φ′iφ

′i−1dx

∫

ei

φ′iφ

′idx

. (4.66)

The known quantities do not occur in the stiffness-matrix, hence for e1 weobtain:

se1 =

∫

e1

φ′1φ

′1dx

, (4.67)


and for e2 and e3:

se2 =

∫

e2

φ′1φ

′1dx

∫

e2

φ′1φ

′2dx

∫

e2

φ′2φ

′1dx

∫

e2

φ′2φ

′2dx

, (4.68)

se3 =

∫

e3

φ′2φ

′2dx

∫

e3

φ′2φ

′3dx

∫

e3

φ′3φ

′2dx

∫

e3

φ′3φ

′3dx

. (4.69)

The matrices se1, se2 and se3 are the element-matrices of elements e1, e2 ande3. We will put these matrices into A0 and add them to obtain A. The positionof the element-matrices is such that se1[1, 1] = A0[1, 1], hence:

A1 =

∫

e1

φ′1φ

′1dx 0 0

0 0 0

0 0 0

. (4.70)

Subsequently we add se2 to A1 such that se2[2, 2] = A0[2, 2]:

A2 =

∫

e1

φ′1φ

′1dx 0 0

0 0 0

0 0 0

+

∫

e2

φ′1φ

′1dx

∫

e2

φ′1φ

′2dx 0

∫

e2

φ′2φ

′1dx

∫

e2

φ′2φ

′2dx 0

0 0 0

=

∫

e1

φ′1φ

′1dx +

∫

e2

φ′1φ

′1dx

∫

e2

φ′1φ

′2dx 0

∫

e2

φ′2φ

′1dx

∫

e2

φ′2φ

′2dx 0

0 0 0

. (4.71)

Subsequently we add se3 to A2 such that se3[2, 2] = A0[3, 3] to obtain A3:

A3 =

∫

e1

φ′1φ

′1dx +

∫

e2

φ′1φ

′1dx

∫

e2

φ′1φ

′2dx 0

∫

e2

φ′2φ

′1dx

∫

e2

φ′2φ

′2dx 0

0 0 0

+

0 0 0

0∫

e3

φ′2φ

′2dx

∫

e3

φ′2φ

′3dx

0∫

e3

φ′3φ

′2dx

∫

e3

φ′3φ

′3dx

=

∫

e1

φ′1φ

′1dx +

∫

e2

φ′1φ

′1dx

∫

e2

φ′1φ

′2dx 0

∫

e2

φ′2φ

′1dx

∫

e2

φ′2φ

′2dx +

∫

e3

φ′2φ

′2dx

∫

e3

φ′2φ

′3dx

0∫

e3

φ′3φ

′2dx

∫

e3

φ′3φ

′3dx

. (4.72)

Now, the stiffness-matrix has been built. The same principle is carried out withthe element-vectors to generate the right-hand side vector. The same principlegives

fei

=

∫

ei

f(x)φi−1(x)dx

∫

ei

f(x)φi(x)dx

. (4.73)

4.8 Numerical integration 42

This is called the element-vector of element ei.

fe1

=

∫

e1

f(x)φ1(x)dx

⇒ f1

=

∫

e1

f(x)φ1(x)dx

0

0

. (4.74)

subsequently

fe2

=

∫

e2

f(x)φ1(x)dx

∫

e2

f(x)φ2(x)dx

(4.75)

⇒ f2

= f1+

∫

e2

f(x)φ1(x)dx

∫

e2

f(x)φ2(x)dx

0

(4.76)

=

∫

e1

f(x)φ1(x)dx +∫

e2

f(x)φ1(x)dx

∫

e2

f(x)φ2(x)dx

0

. (4.77)

Subsequently

fe3

=

∫

e3

f(x)φ2(x)dx

∫

e3

f(x)φ3(x)dx

⇒ f

3=

∫

e1∪e2

f(x)φ1(x)dx

∫

e2∪e3

f(x)φ2(x)dx

∫

e3

f(x)φ3(x)dx

. (4.78)

Now we have the right-handside if we take the essential condition into account

for element e1:

[

−u0

∫

e1

φ′0φ

′1dx

]

at f ′3, to obtain:

f =

∫

e1∪e2

f(x)φ1(x)dx − u0

∫

e1

φ′0φ

′1dx

∫

e2∪e3

f(x)φ2(x)dx

∫

e3

f(x)φ3(x)dx

. (4.79)

The principle of element-matrices and vectors should be seen as a convenientprogramming trick. To tackle non-standard Finite Element problems by theuse of Finite Element software packages, it is sufficient to create the element-matrices and element-vectors.

4.8 Numerical integration

Consider the following weak form

4.9 Error considerations 43

Find u(x), subject to u(0) = u0, such that

−

1∫

0

u′(x)v′(x)dx =

1∫

0

f(x)v(x)dx ∀v(0) = 0. (4.80)

Sometimes, the right-hand side contains a complicated integrand that cannotbe integrated in terms of an elementary function. Since one uses the concept ofelement vectors and element matrices, one has to integrate the functions overan element. Many numerical integration methods are available, such as themidpoint Rule, the Trapeziodal Rule, Simpson Rule and so on. A very com-mon numerical integration rule based on the expression of a function in termsof a linear combination of basis-function is the Newton-Cotes integration Rule.In this section, we will consider Newton-Cotes integration with piecewise linearbasis-functions. This rule is based on the following: Consider a function g(x)to be integrated over an element ei := [xi−1, xi]:

∫

ei

g(x)dx, (4.81)

then we express g(x) as a linear combination of basis-functions with the char-acteristics as mentioned before on the nodes xi−1 and xi:

g(x) = g(xi−1)φi−1(x) + g(xi)φi(x). (4.82)

Then the integration over the interval zero-one, gives:

∫

ei

g(x)dx = g(xi−1)

∫

ei

φi−1(x)dx+g(xi)

∫

ei

φi(x)dx =g(xi−1) + g(xi)

2(xi−xi−1).

(4.83)Note that the above integration formula represents the Trapezoidal Rule fornumerical integration. The same is done with the right-hand of the above weakform:

∫

ei

f(x)φk(x)dx =i∑

j=i−1

f(xj)φk(xj)

∫

ei

φj(x)dx =1

2f(xk)(xk+1−xk−1), (4.84)

for k ∈ i − 1, i, note that φk(xi) = 1 if i = k and else φk(xi) = 0. Thisintegration rule is easily extended to more dimensional problems. This is beyondthe scope of the present course. As an alternative, Gauss integration formulasare used. This is not treated in this course.

4.9 Error considerations

This section is not necessary to understand the finite element method. It isintended for the interested reader. The treatment of the error of the finiteelement solution will not be mathematical, but it will give an idea of what the


error is and what issues are important for the derivation of the error. Supposethat we solve the equation in the following weak form of Poisson’s equation

Find u ∈ H1(Ω), subject to u = g on ∂Ω, such that

∫

Ω ∇u · ∇φdΩ =∫

Ω φfdΩ, for all φ ∈ H1(Ω).

(4.85)

To solve this problem using Galerkin’s method, we approximate the solution byu(x) ≈ uh(x) =

∑nj=1 ujφj(x), where φj(x) = 0 on ∂Ω. Here u(x) represents

the exact solution and uj represents the approximate solution at the nodalpoints. For the approximate solution uh, we have

Find uh ∈ H1h(Ω), subject to uh = g on ∂Ω, such that

∫

Ω ∇uh · ∇φhdΩ =∫

Ω φhfdΩ, for all φh ∈ H1h(Ω),

(4.86)

where H1h(Ω) represents the set of solutions for the approximate solution uh.

Note that H1h(Ω) ⊂ H1(Ω). Let u(x) be the approximate solution with the

exact values of the solution u(x), that is

u(x) =n∑

j=1

u(xj)φj(x).

Note that the difference between the exact solution u(x) and the above solutionu(x) is only determined by the interpolation method that is used. Then it canbe proved (see Braess [1996] for instance) that

∫

Ω||∇(u − uh||

2dΩ ≤

∫

Ω||∇(u − u)||2dΩ. (4.87)

The above integrals represent errors in the energy-norm. The left-hand sideof the above inequality gives the total error of the finite element solution withrespect to the exact solution in the so-called energy norm. This total errorbasically has two sources:

[1] A finite set of basis functions (based on the finite number of meshpoints) ischosen, and hence the summation only concerns a finite number of terms;

[2] Using interpolation functions for the basis functions, this gives an inter-polation error, which depends on the order of the interpolation functionsφj(x).

The right-hand side of the above inequality only concerns the interpolationerror. The above inequality (4.87) is very convenient since it says in the energynorm that the total error is bounded from above by the interpolation error. Thislastmentioned error depends on the polynomial order. If linear elements areused, then it can be demonstrated that the energy norm of the interpolationalerror is of order O(h) where h represents the largest side of the element. Theactual error u−uh is one order higher, that is O(h2). The reader should realize


that many statements in this section have been made while omitting somesubtile mathematical issues.

As a rule of thumb, we use that if the interpolational error is of order O(hp),then the order of the error, that is the difference between the finite elementsolution and the exact solution is of order O(hp+1). It should be mentionedthat this fact only holds if the elements are not degenerate (the vertices of thetriangle are not located on a line in the case of triangular elements).

5

Time dependent problems: numericalmethods

In many cases the mathematical models are ”time-dependent”, by which wemean that their solution depends on time. Typical examples are

∂c

∂t+ q · ∇c = ∆c (convection-diffusion) (5.1)

∂c

∂t− ∆c = 0 (diffusion) (5.2)

∂2c

∂t2= ∆c (wave transmission) (5.3)

∂c

∂t+

∂

∂xf(c) = 0 (Buckley-Leverett in 2-plane flow) (5.4)

In this chapter we will treat some time-integration methods in relation to Fi-nite Diffrences and Finite Elements. Important subjects will be consistency(convergence), accuracy and stability.

5.1 Time-integration methods

First we consider an ODE (Ordinary Differential Equation) to illustrate sometime-integration methods. Subsequently, we will apply these methods to PDE’s.Consider the following problem:

y(0) = y0 t = 0,

y′(t) = f(y, t) t > 0.(5.5)

Let Tend be the end-time of the numerical simulation h =Tend

m, m= number

of timesteps, then we introduce the notation yn = y(tn) = y(nh). We willformulate some classical time-integration methods.

5.2 Accuracy of Time-integration methods 47

[1] Euler’s forward time integration method (explicit):

yn+1 = yn + hf(yn, tn) n ≥ 0 n ∈ 0, . . . ,m − 1 (5.6)

[2] Heun’s (or improved Euler’s) time integration method

yn+1 = yn + hf(yn, tn) (predictor)

yn+1 = yn + h2

[

f(yn, tn) + f(yn+1, tn+1)]

(corrector)(5.7)

[3] Euler’s backward time integration method (implicit)

yn+1 = yn + hf(yn+1, tn+1) (5.8)

(Note that we have to solve a non-linear equation whenever f is non-linearin y)

[4] Crank-Nicholson’s time integration method

yn+1 = yn +h

2

[

f(yn, tn) + f(yn+1, tn+1)]

(5.9)

(Here again a non-linear problem has to be solved when f is nonlinear iny)

We can sum up many more methods for time-integration (see for instance Bur-den and Faires [2001]) methods. These four methods are the most commonand we will analyse them. We also note that the modified Euler method fallswithin the class of the so-called multi-step methods (due to Runge-Kutta).These methods can be adjusted such that always the desired accuracy can beobtained.

5.2 Accuracy of Time-integration methods

We will consider local truncation error. Given, again

y(0) = y0 t = 0,

y′(t) = f(y, t) t > 0,(5.10)

then

y(t + h) = y(t) + hy′(t) +h2

2y′′(t) +

h3

6y′′′(t) + O(h4). (5.11)

Further, let u be the approximated solution and let y(tn) = un, then

u(t + h) = un+1 = yn + hf(yn, tn) (Euler). (5.12)

Since f(yn, tn) = y′(t). Then it follows from (5.11) and (5.12), that

y(t + h) − un+1 =h2

2y′′(t) = O(h2). (5.13)

5.2 Accuracy of Time-integration methods 48

Herewith, the local truncation error is given by:

τn+1(h) =y(t + h) − un+1

h=

h

2y′′(t) = O(h). (5.14)

Hence the order of the local truncation error is h for the Euler-method. This issimilar for the implicit Euler method. We do the same analysis for the accuracyof the modified Euler method. Consider:

un+1 = un + hf(un, tn),

un+1 = un + h2 (f(un, tn) + f(un+1, tn+1)).

(5.15)

Then, assuming un = yn, gives

un+1 = un +h

2f(un, tn) + f(un + hf(un, tn), tn + h)

= yn +h

2f(yn, tn) + f(yn, tn) + hf(yn, tn)

∂f

∂y(yn, tn) + h

∂f

∂t(yn, tn),

(5.16)

this follows from a Taylor expansion. Further:

d2y

dt2=

d

dtf(y(t), t) =

∂f

∂yy′ +

∂f

∂t=

∂f

∂yf +

∂f

∂t. (5.17)

From this follows

un+1 = yn + hf(yn, tn) +h2

2

d2y

dt2(tn) + O(h3). (5.18)

Henceun+1 − yn+1 = O(h3). (5.19)

Herewith, the local truncation error is given by:

τn+1(h) =yn+1 − un+1

h= O(h2). (5.20)

Hence the accuracy is of 2nd order. The truncation error is of 2nd order, whenthe Modified Euler method is used.

To look at the global error, we keep in mind that at each time-step an errorof O(h3), i.e. Error= kh3(k ∈ R), is made (for Modified Euler method). Sincewe integrate over n time-steps the accumulated error is approximately (Globalerror):

k1h3n = k1h

3 Tend

h= k1h

2. (5.21)

Similary for Euler’s method (Global error):

k2h2n = k2h

2 Tend

h= k2h. (5.22)

For Crank-Nicholson’s method, it can be shown by use of a similar method thatTruncation error ∼ h3 and Global error ∼ h2.

5.3 Time-integration of PDE’s 49

5.3 Time-integration of PDE’s

The time integration method for PDE’s is analogous to ODE’s. Suppose thatthe following problem is given:

∂u

∂t= L(u),

u(x, 0) = u0,

with appropraite (BC),

, (5.23)

where L(u) represents a linear space differential operator; example:

L(u) = ∆u (diffusion), or, (5.24)

L(u) = ∆u − q · ∇u (conv.-diff.). (5.25)

and so on... (5.26)

The time discretization part∂u

∂tis done (for instance) by:

∂u

∂t=

un+1i,j − un

i,j

∆t, (5.27)

where uni,j denotes u(xi, yj, t

n). Then we give the following example:

un+1i,j = un

i,j + ∆tL(un) Forward( Explicit) Euler,

un+1i,j = un

i,j + ∆tL(un+1) Backward (Implicit) Euler,

un+1i,j = un

i,j + ∆t2

L(un+1) + L(un)

Crank-Nicholsen,

un+1i,j = un

i,j + ∆tL(un) Modified Euler (predictor-corrector),

un+1i,j = un

i,j + ∆t2

L(un) + L(un+1)

.

(5.28)

The global errors are, respectively O(∆t), O(∆t), O(∆t2) and O(∆t2). Thiscan be shown by the use of similar procedures as in section 5.2. Now we considerthe application to a diffusion equation as our example. We will illustrate the useof Finite Differences and Finite Elements for a 1-dimensional diffusion problemand a 1-dimensional wave equation.

5.3.1 The Heat equation

We consider the discretization of the heat equation

∂u

∂t=

∂2u

∂x2, 0 < x < 1, t > 0

+ Dirichlet boundary conditions + initial condition.

(5.29)

[1] Finite Differences

∂u

∂t=

ui+1 − 2ui + ui−1

∆x2, (5.30)

un+1i − un

i

∆t=

uni+1 − 2un

i + uni−1

∆x2(Forward Euler), (5.31)

un+1i − un

i

∆t=

un+1i+1 − 2un+1

i + un+1i−1

∆x2(Backward Euler). (5.32)


[2] Finite Elements∂u

∂t=

∂2u

∂x2. (5.33)

Weak form:∫

Ω

∂u

∂tvdx =

∫

Ω

∂2u

∂x2vdx = −

∫

Ω

∂u

∂x

∂v

∂xdx, (5.34)

where we took v |∂Ω= 0. Find u, subject to u(0) = u0, u(1) = u1, suchthat

1∫

0

∂u

∂tvdx = −

1∫

0

∂u

∂x

∂v

∂xdx ∀v(0) = 0 = v(1). (5.35)

Provided that the above integral exist. Further take basis-functions (piece-wise linear), φi(0) = 0 = φi(1) for i ∈ 1, . . . ,m − 1,

u(x, t) =

m−1∑

j=1

[cj(t)φj(x)] + u0φ0(x) + u1φm(x). (5.36)

Take for simplicity u0 = 0 = u1 then

m−1∑

j=1

c′j(t)

1∫

0

φj(x)φi(x)dx = −m−1∑

j=1

cj(t)

1∫

0

φ′j(x)φ′

i(x)dx i ∈ 1, . . . ,m−1.

(5.37)For forward Euler, one obtains:

m−1∑

j=1

cn+1j − cn

j

∆t

1∫

0

φj(x)φi(x)dx = −m−1∑

j=1

cnj

1∫

0

φ′j(x)φ′

i(x)dx i ∈ 1, . . . ,m−1,

(5.38)

where we call Mij =1∫

0

φi(x)φj(x)dx the entries of the mass-matrix and

Sij =1∫

0

φ′i(x)φ′

j(x)dx the entries of the stiffnes-matrix. Elementwise, we

obtain

cn+1i−1 − cn

i−1

∆t

∫

ei

φiφi−1dx +cn+1i − cn

i

∆t

∫

ei∪ei+1

(φi)2dx +

cn+1i+1 − cn

i+1

∆t

∫

ei+1

φiφi+1dx

= −cni−1

∫

ei

φ′iφ

′i−1dx − cn

i

∫

ei∪ei+1

φ′iφ

′idx − cn

i+1

∫

ei+1

φ′iφ

′i+1dx for i ∈ 1, ..,m − 1.

(5.39)

All the integrals can be computed. The discretization method for FEM looksvery different. For the case of an equidistant grid, it can be shown that the


result becomes identical to FVM and FDM. For backward Euler, one obtainssimilarly:

cn+1i−1 − cn

i−1

∆t

∫

ei

φiφi−1dx +cn+1i − cn

i

∆t

∫

ei∪ei+1

(φi)2dx +

cn+1i+1 − cn

i+1

∆t

∫

ei+1

φiφi+1dx

= −cn+1i−1

∫

ei

φ′iφ

′i−1dx − cn+1

i

∫

ei∪ei+1

φ′iφ

′idx − cn+1

i+1

∫

ei+1

φ′iφ

′i+1dx

for i ∈ 1, ..,m − 1. (5.40)

All these expressions can be adjusted to Crank-Nicholson or a two-step method(Runge-Kutta 2 or Modified Euler). Later we will consider the stability ofnumerical methods.

We remark further that the above Finite Element dicretization of the heatproblem can be written in matrix-form as

Mdc

dt= Sc, (5.41)

where Mij =1∫

0

φi(x)φj(x)dx, Sij =1∫

0

φ′i(x)φ′

j(x)dx and c = [c1 . . . cn]T . Of

course, the Euler, modified Euler etc. can be written similarly as before. As anexample the Euler-forward time integration method can be written by:

Mcn+1 = Mcn + ∆tScn. (5.42)

For the case that M is not a diagonal matrix, a linear system of equations hasto be solved also for the Euler-forward method.

Exercise 5.3.1. Write down the corresponding time-integration methods forthe backward Euler, Modified Euler and Crank-Nicholson methods.

5.3.2 The wave equation

We consider the discretization of the wave equation

∂2u

∂t2= c2 ∂2u

∂x2, 0 < x < 1, t > 0,

+ Dirichlet boundary conditions + initial conditions for u and ut.

(5.43)

Here c represents the given wave-speed, which is assumed to be constant in thetext. The weak form of the above equation is obtained after multiplication bya test-function, to obtain

∫

Ω

∂2u

∂t2vdx = c2

∫

Ω

∂2u

∂x2vdx = −c2

∫

Ω

∂u

∂x

∂v

∂xdx, (5.44)


where we took v |∂Ω= 0. Find u, subject to u(0) = u0, u(1) = u1, such that

1∫

0

∂2u

∂t2vdx = −c2

1∫

0

∂u

∂x

∂v

∂xdx ∀v(0) = 0 = v(1). (5.45)

Provided that the above integral exist. Further take basis-functions (piecewiselinear), φi(0) = 0 = φi(1) for i ∈ 1, . . . ,m − 1,

u(x, t) =

m−1∑

j=1

[cj(t)φj(x)] + u0φ0(x) + u1φm(x). (5.46)

Take for simplicity u0 = 0 = u1 then

m−1∑

j=1

c′′j (t)

1∫

0

φj(x)φi(x)dx = −c2m−1∑

j=1

cj(t)

1∫

0

φ′j(x)φ′

i(x)dx i ∈ 1, . . . ,m − 1.

(5.47)

where we call Mij =1∫

0

φi(x)φj(x)dx the entries of the mass-matrix and Sij =

1∫

0

φ′i(x)φ′

j(x)dx the entries of the stiffnes-matrix. Hence, we obtain

Md2c

dt2= c2Sc, (5.48)

One possibility to integrate the above equation in time is to write it as systemof first-order differential equations in time for c and w:

dc

dt= w,

Mdw

dt2= c2Sc,

with initial conditions for c and w.

(5.49)

Now, we can apply the Forward Euler method to the above problem:

cn+1 = cn + ∆twn;

Mwn+1 = Mwn + ∆tc2Scn.

(5.50)

Exercise 5.3.2. Write down the corresponding time-integration methods forthe backward Euler, Modified Euler and Crank-Nicholson methods.

It can be shown that the numerical integration of the above equation gives riseto dissipation cn → 0 as n → ∞). The amount of dissipation can be de-creased when a Runge-Kutta method is used. Further one can use a directtime-integration of the second-order system of differential equations. This isnot treated any further.

5.4 Stability analysis 53

5.4 Stability analysis

In general after discretization we obtain a system of ordinary differential equa-tion in the form

Mdu

dt= Su ⇔

du

dt= M−1Su. (5.51)

For the stability the eigenvalues of the above matrix M−1S are crucial. First,for the analytical asymptotic stability we require

λ(M−1S) < 0, and if λ /∈ R we have ℜλ(M−1S) < 0. (5.52)

Exercise 5.4.1. Derive stability criteria for the eigenvalues of M−1S for theForward Euler, Modified Euler, Backward Euler and Trapezoidal time integra-tion methods.

Hence, it is crucially important to have some knowledge on the eigenvalues ofthe matrix M−1S. For large matrices it is not easy to compute the eigenval-ues. Fortunately, Gershgorin’s Theorem gives a often very usefull estimate ofthe eigenvalues. For many cases the mass matrix M is diagonal as a result ofnumerical integration by use of the Rule of Newton-Cotes. This is commonlycalled lumping.

Exercise 5.4.2. Show by the numerical integration of Newton-Cotes that

mji =

∫ 1

0φi(x)φj(x)dx =

xi+1 − xi−1

2, if j = i,

0, else.

(5.53)

For this case the Theorem reads as follows:

Theorem 5.4.1. Let M be diagonal, then, for all eigenvalues λ of M−1S holds:

|λ| ≤ supk

1

|mkk|

n∑

i=1

|ski|. (5.54)

Note that the eigenvalues may be complex if S is not symmetric. Then the areasin which the eigenvalues are allowed to be in, consist of circles in the complexplane.

Proof 5.4.1. Let λ be an eigenvalue of the generalized eigenvalue problem withcorrespoding eigenvector v, then,

Sv = λMv. (5.55)

5.4 Stability analysis 54

In other words, this becomes for each component k:

n∑

i=1

skivi = λmkkvk. (5.56)

Let vj be the component of v with the largest modulus, then, we have for thisindex j:

λ =1

mjj

n∑

i=1

sjivi

vj, (5.57)

and since |vi/vj | ≤ 1, we get

|λ| ≤1

|mkk|

n∑

i=1

|ski|. (5.58)

This proves the assertion.

We illustrate the use of Gershgorin’s Theorem by the following example.

Example: Suppose we use Finite Differences, then M = I. Let h be thestepsize and ∆t be the time-step. Further in one dimension, we have sii =−2/h2 and sii−1 = 1/h2 = sii+1. From this, we obtain λ ≤ 4/h2. Hence weobtain for the Forward Euler method: ∆t ≤ h2/2.

The analysis of stability of time-integration methods for Partial DifferentialEquations can also be done by the use of the Von Neumann analysis, which isbased on Fourier analysis. This is omitted in the present course.

6

Numerical linear algebra

In this part we will consider some basic methods to solve the matrix-vectorequations that arise from the various types of discretizations. For small systems,one preferably uses a so-called “direct” method to solve the equation. Forlarge, sparse (i.e. with mainly zeroes in the matrix and non-zero entries in thevicinity of the main diagonal) systems/matrices, one better relies on iterativemethods. The treatment given here is far from complete. Before we treat somemethods, we will review some basic linear algebra concepts, which are essentialfor understanding.

6.1 Basics

Given a matrix A and vector x:

A =

a11 a12 . . . a1n

a21

...

an1 . . . . . . ann

, x =

x1

x2

...

xn

(6.1)

then we can see A as a row of columns: A = (a1a2 . . . an), where

ai =

a1i

a2i

...

ani

, i ∈ 1, .., n. (6.2)

6.1 Basics 56

The product of A and x can be written by

Ax =(

a1a2 . . . an

)

x1

x2

...

xn

:= a1x1 + a2x2 + · · · + anxn

=

a11

a21

...

an1

x1 +

a12

a22

...

an2

x2 + · · · +

a1n

a2n

...

ann

xn

=

a11x1 + a12x2 + · · · + a1nxn

a21x1 + a22x2 + · · · + a2nxn

...

an1x1 + an2x2 + · · · + annxn

. (6.3)

6.1.1 Eigenvalues and eigenvectors of A

The eigenvalues and eigenvectors are defined by the solution of the followingproblem:

[

Find λ, v such that

Av = λv for v 6= 0,(6.4)

λ is called an eigenvalue and v is called an eigenvector. This implies thatAv − λv = 0 for v 6= 0. Let I = diag(1 . . . 1) be the identity matrix then itfollows that

(A − λI)v = 0 for v 6= 0. (6.5)

This implies that A−λI is singular and hence that its determinant is zero, i.e.

det(A − λI) = 0. (6.6)

The above equation is referred to as the characteristic equation, which definesthe eigenvalues (λ).

Example: Given

A =

(

2 −1

−1 2

)

. (6.7)

Eigenvalues

det(A − λI) = 0 ⇒

∣

∣

∣

∣

∣

2 − λ −1

−1 2 − λ

∣

∣

∣

∣

∣

= 0 (6.8)

⇔ (2 − λ)2 − 1 = 0 ⇔ 2 − λ = ±1 ⇒ (6.9)

λ = 3 ∨ λ = 1. (6.10)

6.1 Basics 57

Eigenvectors:

Av = λv, (6.11)

λ = 3 :

(

−1 −1

−1 −1

)(

v(1)

v(2)

)

=

(

0

0

)

⇒ v =

(

1

−1

)

, (6.12)

λ = 1 :

(

1 −1

−1 1

)(

v(1)

v(2)

)

=

(

0

0

)

⇒ v =

(

1

1

)

. (6.13)

6.1.2 Diagonalization of A

Suppose that λ1, . . . , λn are the eigenvalues of A with eigenvectors v1, . . . , vn

and let P =(v1 v2 . . . vn), then

AP = A(v1 v2 . . . vn) = (Av1 Av2 . . . Avn) = (λ1v1 . . . λnvn). (6.14)

Further let D=diag(λ1 λ2 . . . λn) then

PD = (v1 v2 . . . vn)

λ1 ∅. . .

∅ λn

= (λ1v1 λ2v2 . . . λnvn). (6.15)

Hence AP = PD . Note that we assumed that for each λi, there exists a vi.

Hence A = PDP−1 →diagonalization. Suppose that A is symmetric, aij =aji, AT = A, then

A = PDP−1, (6.16)

AT = (PDP−1)T = P−TDP T . (6.17)

Since A = AT , we obtain that P T = P−1 satisfies! Hence when A = AT , then

A = PDP T .

6.1.3 Eigenvalues of a symmetric matrix

Theorem 6.1.1. The eigenvalues of a real-valued symmetric matrix are real.Further its eigenvectors of different eigenvalues are orthogonal.

Before we prove this assertion, we introduce the concept of inner-product andorthogonality. Given two vectors in Cn, then we define its inner product by

〈v1, v2〉 = vT1 v2, (6.18)

6.2 Solution of systems of linear equations 58

where v1 is the complex conjugate of v1. Consequently, when v1, v2 ∈ Rn,A ∈ Rn×n, then

〈Av1, v2〉 = (Av1)T v2 = vT

1 AT v2 =⟨

v1, AT v2

⟩

. (6.19)

Hence 〈Av1, v2〉 =⟨

v1, AT v2

⟩

. Now we consider the proof of Theorem 6.1.1:

Proof 6.1.1. We compute

〈Av1, v1〉 = 〈λ1v1, v1〉 = λ1 〈v1, v1〉 . (6.20)

Further〈v,Av1〉 = 〈v1, λ1v1〉 = λ1 〈v1, v1〉 . (6.21)

Since A = AT , it follows that 〈Av1, v1〉 = 〈v1, Av1〉, and hence the expressions(6.20) and (6.21) should be equal. Hence λ1 = λ1 and hence λ1 ∈ R. This holdsfor all eigenvalues analogously. (since Avi = λivi). Two vectors are orthogonalif and only if 〈v1, v2〉 = 0. Compute

〈Av1, v2〉 = λ1 〈v1, v2〉 , (6.22)

since λ1 ∈ R. Further〈v1, Av2〉 = λ2 〈v1, v2〉 . (6.23)

Since A = AT we have 〈Av1, v2〉 = 〈v1, Av2〉 and hence

λ1 〈v1, v2〉 = λ2 〈v1, v2〉

⇔ 〈v1, v2〉 = 0. (6.24)

This implies that v1 and v2 are orthogonal.

6.2 Solution of systems of linear equations

6.2.1 The LU -decomposition

Consider the following system of equations:

a11x1 + a12x2 + · · · + a1nxn = b1,

a21x1 + a22x2 + · · · + a2nxn = b2,...

an1x1 + an2x2 + · · · + annxn = bn.

(6.25)

We can see the above system as a matrix equation

a11 a12 . . . a1n

a21 a22 . . . a2n

...

an1 an2 . . . ann

x1

x2

...

xn

=

b1

b2

...

bn

, (6.26)


Ax = b . (6.27)

Direct methods are Gauss elimination, LU -decomposition. LU -decompositionis used in many practical situations: Therefore, we consider the following ex-ample: Let

A =

(

a11 a12

a21 a22

)

−a21a11

↓∼

(

a11 a12

0 a22 −a21a11

a12

)

= U. (6.28)

Further let

L =

(

1 0a21a11

1

)

. (6.29)

We calculate the product of LU :

LU =

(

1 0a21a11

1

)(

a11 a12

0 a22 −a21a11

a12

)

(6.30)

=

(

a11 a12a21a11

a11 a12a21a11

+ a22 −a21a11

a12

)

=

(

a11 a12

a21 a22

)

= A. (6.31)

Hence

A = LU

L : lower matrix (triangular).

U : upper triangular matrix.(6.32)

Ax = b ⇔ LUx = b, (6.33)

⇒

Ls = b,

Ux = s,(6.34)

solve s and x by use of these two steps in (6.34). For larger matrices this readsas

L =

1

l21 1 ∅

l31 l32 1...

. . .. . .

ln1 . . . . . . lnn−1 1

, U =

a11 a12 . . . . . . a1n

a22 . . . . . . a2n

a33 . . . a3n

∅. . .

...

ann

(6.35)

Ls = b ⇒

s1 = b1

s2 = b2 − l21s1 = b2 − l21b1

etc.

(6.36)

and xn =sn

ann, etc. This gives a gain in the efficiency. For one-dimensional

problems, this is a nice method. For sparse matrices for 3D geometries this isnot efficient due to ”fillin”, which is the loss of the sparse band structure of thematrix.


The LU -decomposition A = LU can be obtained by Gauss-elimination:

A =

a11 a12 . . . a1n

a21 a22 . . . a2n

...

an1 an2 . . . ann

−a21a11

−a31a11

. . . −an1a11

↓ | |

↓ |

↓

(6.37)

∼

a11 a12 . . . a1n

0 a(1)22 . . . a

(1)2n

...

0 a(1)n2 . . . a

(1)nn

= U (1), (6.38)

where a(1)22 = a22 −

a21a11

a11 etc, and a(1)n2 = an2 −

an1a11

a11 etc.

L(1) =

1a21a11

1 ∅... ∅

. . .an1a11

∅ . . . 1

. (6.39)

We can check that L(1)U (1) = A. We repeat this on U (1),

U (1) =

a11 a12 . . . a1n

a(1)22 . . . a

(1)2n

∅ a(1)22 . . . a

(1)2n

...

a(1)n2 . . . a

(1)nn

−a(1)32

a(1)22

. . . −a(1)n2

a(1)22

↓ |

|

↓

(6.40)

∼

a11 a12 . . . . . . a1n

0 a(1)22 . . . . . . a

(1)2n

0 0 a(2)33 . . . a

(2)3n

∅...

...

a(2)n3 . . . a

(2)nn

= U (2) (6.41)

L(2) =

1a21a11

1 ∅

a31a11

a(1)32

a(1)22

1

......

. . .

an1a11

a(1)n2

a(1)nn

1

. (6.42)

We can check that A = L(2)U (2). This procedure repeated until U (n) is anupper matrix (triangle). Hence, n times!


An alternative is Crout’s scheme:

A = LU, (6.43)

suppose that

A =

(

A ac

aTr ann

)

, (6.44)

and that A = LU is known, then

L =

(

L ∅

lTr 1

)

and U =

(

U uc

∅ unn

)

. (6.45)

Then

A = LU ⇒

(

A ac

aTr ann

)

=

(

L ∅

lTr 1

)(

U uc

∅ unn

)

. (6.46)

Evaluation of the products of these block matrices, gives:

A = LU

aTr = lTr U ⇒ UT lr = ar

ac = Luc

ann = lTr uc + unn ⇒ unn = ann − lTr uc

(6.47)

Here we have to determine lTr , uc and unn.


A =

2 −1 0

−1 2 −1

0 −1 2

. (6.48)

Give the LU -decomposition of A.

A disadvantage is the fill-in that takes place for more-dimensional problems:

A =

∅

∅

∅

∅

∼ · · · ∼

extra ∅

non-

∅ zero’s

∅

= U

(6.49)and

L =

extra ∅

non-

∅ zero’s

(6.50)

Especially when A is a discretization matrix of a 3D-model then this will destroythe efficiency. For these problems one takes refuge to iterative solvers.


6.2.2 Basic, classical iterative solvers

We consider again

a11x1 + a12x2 + · · · + a1nxn = b1,

a21x1 + a22x2 + · · · + a2nxn = b2,...

an1x1 + an2x2 + · · · + annxn = bn.

(6.51)

Ax = b, consider the sequence of xk = (xk1 xk

2 . . . xkn)T were xk is the solution

after k iterations. Then, Jacobi-iterations work as follows:

a11xk+11 = b1 − a12x

k2 − a13x

k3 · · · − a1nxk

n,

a22xk+12 = b2 − a21x

k1 − a23x

k3 · · · − a2nxk

n,...

annxk+1n = bn − an1x

k1 − an2x

k2 · · · − annxk

n.

(6.52)

In other words, when we introduce the following notation:

• D=diag(a11, a22, . . . , ann),

• Lij = −Aij when i > j, Lij = 0 when i ≤ j (lower),

• Uij = −Aij when i < j, Uij = 0 when i ≥ j (upper),

then A = D − L − U . Jacobi-iteration then works as follows

Find x (initial choice/estimate)

Dxk+1 = b + (L + U)xk, k ∈ 0, 1, 2, . . . .(6.53)

The Jacobi-iterations represent a classical iterative solution method. It con-verges however very slowly.

Looking at equations (system (6.52), we see that after xk+11 is determined,

it could be substituted into the equation for xk+12 . Subsequently, the values

of xk+11 and xk+1

2 can be substituted into the equation for xk+13 , and so on.

This will give, on intuitive grounds, a faster convergence behaviour. This is theiteration method due to Gauss-Seidel. It runs as follows:

a11xk+11 = b1 − a12x

k2 − a13x

k3 · · · − a1nxk

n,

a22xk+12 = b2 − a21x

k+11 − a23x

k3 · · · − a2nxk

n,

a33xk+13 = b3 − a31x

k+11 − a32x

k+12 · · · − a3nxk

n,...

annxk+1n = bn − an1x

k+12 − an2x

k+12 · · · − annxk+1

n .

(6.54)

This implies in vector-form:

Choose x

(D − L)xk+1 = Uxk + b.(6.55)

6.3 Convergence properties of iterative methods 63

This is known as Gauss-Seidel’s iteration method. Convergence is faster thanthe convergence of Jacobi’s method.


b =(

1 1 1 1 1 1 1 1 1)T

(6.56)

and

A =

4 −1 0 −1 0 0 0 0 0

−1 4 −1 0 −1 0 0 0 0

0 −1 4 0 0 −1 0 0 0

−1 0 0 4 −1 0 −1 0 0

0 −1 0 −1 4 −1 0 −1 0

0 0 −1 0 −1 4 0 0 −1

0 0 0 −1 0 0 4 −1 0

0 0 0 0 −1 0 −1 4 −1

0 0 0 0 0 −1 0 −1 4

(6.57)

Give the matrices D, L and U for the Gauss-Seidel method.

Now we will analyze some convergence properties.

6.3 Convergence properties of iterative methods

The Jacobi and Gauss-Seidel iterations can be written in the following form:

(D − L)xk+1 = kxk + b ⇒ xk+1 = (D − L)−1Uxk + (D − L)−1, (6.58)

provided that det(D − L) 6= 0. In other words when we put M := (D −L)−1Uand f := (D − U)−1b, then

xk+1 = Mxk + f . (6.59)

This is the most basic iterative scheme. We will analyze its convergence. There-fore, we need to prove the following theorem:

Theorem 6.3.1. Let xk ∈ Rn, M ∈ Rn×n and f ∈ Rn.

The sequence xk+1 = Mxk + f converges to a limit if and only if | λ1 |< 1,where λ1 is the eigenvalue of matrix M with the largest absolute value.

Before we ”prove” this assertion we remind that we mean convergence if thereexists a vector x for which xk → x as k → ∞. In other words xk has a limit.Mathematicians are rather formal about convergence (and that’s their problemand right), and they will use the following definition in general:


Definition 6.3.2. A sequence of vectors is called convergent to x if there existsε > 0 and N ∈ N such that

| xk − x |< ε for all n > N(= N(ε)). (6.60)

We then have xk → x as n → ∞.

The proof that we present here, will not be very formal. We sketch the prooffor diagonalizable matrices only, i.e. There exists P , Λ = diag(λ1, . . . , λn) suchthat A = PDP−1, where the columns of P represent eigenvectors, see Section6.1. The proof can be generalized using Jordan decompositions with general-ized eigenvectors for non-diagonalizable matrices. An alternative proof is basedon writing the start-vector as a linear combination of the eigenvectors of thematrix M . These approaches will be omitted.

Proof 6.3.1. (Proof of Theorem 6.3.1): We assume that M = PΛP−1, whereΛ = diag(λ1, . . . , λn) with λi representing the eigenvalue of M and P =(v1, . . . , vn) is produced by the use of eigenvectors of M (see section 6.1). Then,

xk+1 = PΛP−1xk + f, (6.61)

or equivalentlyP−1xk+1 = ΛP−1xk + P−1f, (6.62)

where P−1xk+1 = yk+1, ΛP−1xk = Λyk and P−1f = f . Introduction of yk :=

P−1xk and f = P−1f , then gives:

yk+1 = Λyk + f . (6.63)

Suppose that this limit exists, i.e. yk → y as k → ∞, then

y = Λy + f holds for the exact limit, and

yk+1 = Λyk + f

−

y − yk+1 = Λ(y − yk) (6.64)

where y − yk+1 is the error after k + 1 iterations and y − yk is the error after

k iterations. Component-wise we see yi − yk+1i = λi(yi − yk

i ), hence

| yi − yk+1i |≤| λi | · | yi − yk

i | . (6.65)

We see that | λi |< 1 implies | y − yk+1 |≤| y − yk |⇔ yk → y. Further, we see

that if yk → y then

| yi − yk+1i |≤| yi − yk

i | (i.e. convergence), (6.66)

implies | λi |< 1. This proves the Theorem.


Consequence: The iterative methods of Jacobi and Gauss-Seidel convergeif and only if

| λ(D−1(L + U)) |< 1,

| λ((D − L)−1U) |< 1.(6.67)

We see from the above that it is important to know the eigenvalues. However,for large matrices the computation of the eigenvalues is expensive. We will pro-vide some Theorems on the estimation of the eigenvalues. First we will showthat Gauss-Seidel converges if A is a symmetric and positive definite matrix.First we need the definition of positive definitity:

Definition 6.3.3. A real-valued symmetric matrix (A = AT ) is positive definiteif and only if

〈x,Ax〉 = xT Ax > 0 ∀x ∈ Rn, x 6= 0. (6.68)

We remark that A is positive definite (A = AT ) if and only if λ(A) > 0 forall its eigenvalues. Now we will state a theorem for convergence of Gauss-Seidelfor a symmetric positive definite matrix A.

Theorem 6.3.4. If the matrix A is symmetric positive definite (SPD) then,the Gauss-Seidel process converges.

Proof 6.3.2. (Proof of Theorem 6.3.4): Let A be symmetric A = AT , then

A = D − L − LT (LT = U). (6.69)

Further we have to show that | λ((D − L)−1LT ) |< 1. Let λ and v be aneigenvalue and an eigenvector of (D − L)−1LT , then

(D − L)−1LT v = λv, (6.70)

and(D − L)−1LT v = λv. (6.71)

This impliesLT v = λ(D − L)v, (6.72)

andLT v = λ(D − L)v. (6.73)

Further we will compute 〈v,Av〉 by the use of the two following possibilities:

[1]

〈v,Av〉 = vT Av = vT (D − L − LT )v = vT (D − L)v − vT LT v

= vT (D − L)v − vT λ(D − L)v

= (1 − λ)vT (D − L)v. (6.74)


[2]

〈v,Av〉 = vT Av = vT (D − LT )v − vT Lv = vT (D − LT )v − (LT v)T v

= vT (D − LT )v − (λ(D − L)v)T v

= vT (D − LT )v − λvT (D − LT )v

= (1 − λ)vT (D − LT )v. (6.75)

The above two relations imply:

1

1 − λ〈v,Av〉 = vT (D − L)v

1

1 − λ〈v,Av〉 = vT (D − LT )v

+(

1

1 − λ+

1

1 − λ

)

〈v,Av〉 = 〈v,Dv〉 + 〈v,Av〉 (6.76)

where 〈v,Av〉 > 0, 〈v,Dv〉 > 0 (this can be shown by use of Gershgorin’sTheorem, to be explained later) and 〈v,Av〉 > 0. From this follows

1

1 − λ+

1

1 − λ> 1. (6.77)

Hence

1 − λ + 1 − λ

(1 − λ)(1 − λ)> 1

⇔ 2 − 2Reλ > 1 − (λ + λ)+ | λ |2

⇔ 2 − 2Reλ > 1 − 2Reλ+ | λ |2

⇔ | λ |2< 1 . (6.78)

This implies that | λ((D − L)−1LT ) |< 1 and hence due to Theorem 6.3.1 wehave convergence for Gauss-Seidel.

Now we will derive a theorem, due to the Russian mathematician Gersh-gorin, for the estimation of the eigenvalues of A.

Theorem 6.3.5. Let A ∈ Rn×n with eigenvalues λ1, λ2, . . . , λn then

| λ − aii |≤n∑

j = i

j 6= i

| aij | for all i ∈ 1, . . . , n. (6.79)

The above Theorem provides a bound in which the eigenvalues should be lo-cated. It can be used for the derivation of stability of a numerical method.Let’s prove the theorem!


Proof 6.3.3. (Proof of Theorem 6.3.5): Let λ be an eigenvalue with eigen-vector v, v = (v1, v2, . . . , vn)T , Av = λv, note A = (a1 a2 . . . an), thenAv = (a1 . . . an)v = a1v

1 + a2v2 + · · · + anvn = λv. Consider the ith com-

ponent, which is the largest component of v:

ai1v1 + ai2v

2 + · · · + ainvn = λvi

⇒ (λ − aii)vi = ai1v

1 + ai2v2 + · · · + aii−11v

i−1 + aii+1vi+1 + · · · + ainvn.

(6.80)

Hence:

(λ − aii)vi =

n∑

j = i

j 6= i

aijvj

⇒ | λ − aii |= |n∑

j = i

j 6= i

aij |vj

vi| ≤

n∑

j = i

j 6= i

| aijvj

vi|

≤n∑

j = i

j 6= i

| aij | (since | vj |≤| vi |). (6.81)

This holds for all eigenvalues and hence the theorem is proven.

A consequence of the above theorem is that if a symmetric matrix withpositive elements on the main diagonal is diagonally dominant, i.e.

aii ≥∑

j = i

j 6= i

| aij |> 0), (6.82)

then the matrix is positive (semi) definite. This is due to

| λ − aii | ≤∑

j 6=i

| aij | (6.83)

−∑

j 6=i

| aij |≤ λ − aii ≤∑

j 6=i

| aij | (6.84)

0 ≤ aii −∑

j 6=i

| aij |≤ λ ≤ aii +∑

j 6=i

| aij | (6.85)

Hence λ > 0 and hence A is positive (semi) definite. We, then, proved thefollowing Theorem:

Theorem 6.3.6. Let A = AT ∈ Rn×n and aii > 0, i ∈ i, . . . , n and let

aii >

n∑

j = i

j 6= i

| aij |, then A is positive definite.

6.4 Gradient methods for Ax = b 68

6.4 Gradient methods for Ax = b

The schemes of Gauss-Seidel and Jacobi are classical schemes. A completelydifferent, but fast converging, class is the so-called Krylov-subspace class ofiterative solution methods for linear equations. The mathematical theory is toocomplicated for treatment here. Further, there are numerous choices for Krylovsubspace methods, which will not be treated here. A nice feature of the Krylovmethods is their suitability to large systems and the fact that they only consistof matrix-vector multiplications, which is efficient when the matrix A is sparse.In this section we will only consider a gradient method which is one of the mostimportant principles on which the conjugate gradient method is based. Theclass of gradient methods is based on the minimization of a functional in whichthe matrix A is included. An important condition here is that A is symmetricand positive definite (A is SPD). Consider the following functional J(x):

J(x) =1

2〈x,Ax〉 − 〈x, b〉 , x ∈ Rn. (6.86)

Now, we consider minimization of the above functional:

Find y ∈ Rn such that

J(y) ≤ J(x) for all x ∈ Rn. (6.87)

Now we will show that this minimum co-incides with the solution of Ay = b,where y should be determined. Let x = y + εv, then

J(y) ≤ J(y + εv) ∀v ∈ Rn (6.88)

In other wordsd

dεJ(y + εv) |ε=0= 0 ∀v ∈ Rn. (6.89)

J(y + εv) =1

2

⟨

y + εv,A(y + εv)⟩

−⟨

y + εv, b⟩

=1

2

⟨

y,Ay⟩

+ε

2

⟨

y,Av⟩

+ε

2

⟨

v,Ay⟩

+ε2

2〈v,Av〉 −

⟨

y, b⟩

− ε 〈v, b〉 . (6.90)

Then,d

dεJ(y + εv) |ε=0=

⟨

v,Ay⟩

− 〈v, b〉 = 0, (6.91)

note that⟨

y,Av⟩

=⟨

Ay, v⟩

=⟨

v,Ay⟩

, (6.92)

because AT = A and v, y ∈ Rn, A ∈ Rn×n. Hence the minimization problemtransforms into:


Find y ∈ Rn such that

⟨

v,Ay − b⟩

= 0 ∀v ∈ Rn. (6.93)

This implies that Ay − b = 0 ⇔ Ay = b. So the solution of (6.87) co-incideswith the solution of (6.93) and of:

Ay = b with A SPD, y ∈ Rn. (6.94)

This is the principle on which (conjugate) gradient method are based. Now wetreat the simplest, most basic gradient method that is used to solve (6.87). We,therefore, keep in mind that ∇J(x) gives the direction of the steepest inclineand hence −∇J(x) gives the steepest descent. Now the gradient of J(x) can becomputed by evaluation of all the partial derivatives:

−∇J(x) =

− ∂J∂x1

(x)

− ∂J∂x2

(x)...

− ∂J∂xn

(x)

. (6.95)

This will give −∇J(x) = b − Ax =: r. b − Ax is commonly referred to as theresidual. Suppose that xk represent the set of iterates, then we define:

rk = b − Axk = −∇J(xk). (6.96)

The vector rk gives the direction of the steepest descent. Now we try to walkfrom xk over the direction of rk such that we find xk+1 which minimizes J(x)over this direction: in other wordsFind αk ∈ R such that

J(xk + αkrk) ≤ J(xk + βkrk) ∀β ∈ R. (6.97)

Then, we set xk+1 = xk + αkrk. This αk should be determined and then thisprocess is executed recursively. This gives find αk from

d

dαk

J(xk + αkrk) = 0 ⇒ αk =〈rk, rk〉

〈rk, Ark〉. (6.98)

Now we are able to construct the most basic gradient method. Choose x0,r0 = b − Ax0,

do k = 1 : n

αk−1 =

⟨

rk−1, rk−1

⟩

⟨

rk−1, Ark−1

⟩

xk = xk−1 + αk−1rk−1

rk = b − Axk = rk−1 − αk−1Ark−1

end do


Of course, in practice the do-loop is replaced by a while loop until convergenceis obtained, i.e.

‖ rk ‖< ε. (6.99)

The above algorithm is the basis for the more commonly used conjugate gra-dient method. The CG-method has the important different (from the gradientmethod) characteristic that minimization takes place in the so-called A-norm:

‖ v ‖A:=√

〈v,Av〉. (6.100)

Minimization in the A-norm is natural since the solution Ay = b has to befound where b (i.e. Ay) is known. CG tries to find

min

xk ∈ K‖ y − xk ‖A, (6.101)

where y is exact and xk is the numerical solution in a subspace of Rn. We referto Golub and Van Loan [1996] for a further (algorithmic and mathematical)treatment of the CG-method.

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Introduction into Finite Elementsta.twi.tudelft.nl/users/vermolen/wi3098/wi3098.pdf · The notes provide an introduction into the Finite Element method applied to Partial Diﬀerential

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