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An Introduction to the Mathematical

Foundations of the Finite Element Method

Vitoriano Ruas

October 23, 2002

ii

Aos meus bons amigos da

Universidade Federal Fluminense

Contents

Preamble 1

1 Motivation 3

2 Some Concepts of Functional Analysis 11

2.1 Normed Functions Spaces . . . . . . . . . . . . . . . . . . . . . . 11

2.2 Inner Product Spaces . . . . . . . . . . . . . . . . . . . . . . . . . 13

2.3 Closed Sets in Normed Spaces . . . . . . . . . . . . . . . . . . . . 18

2.4 Complete Normed Spaces . . . . . . . . . . . . . . . . . . . . . . 23

2.5 Orthogonal Projections . . . . . . . . . . . . . . . . . . . . . . . . 29

2.6 Bounded Linear Functionals . . . . . . . . . . . . . . . . . . . . . 31

3 The Finite Element Method in One Dimension Space 37

3.1 An Abstract Linear Variational Problem . . . . . . . . . . . . . . 37

3.2 The Ritz Approximation Method . . . . . . . . . . . . . . . . . . 42

3.3 Piecewise Linear Finite Element Approximations . . . . . . . . . . 45

3.4 More General Cases and Higher Order Methods . . . . . . . . . . 52

4 Extension to Multi-dimensional Problems 59

4.1 Sobolev SpacesHm(Ω) . . . . . . . . . . . . . . . . . . . . . . . . 59

4.2 Green"s Formulae . . . . . . . . . . . . . . . . . . . . . . . . . . . 62

4.3 Model Variational Problems in N-dimensional Spaces . . . . . . . 64

4.4 Lagrange Finite Element Solution Methods . . . . . . . . . . . . . 67

References 73

iv Contents

Preamble

The finite element method is nowadays a widespread tool to solve Engineering problems governed by differential equations of various types. Since it was intro- duced its mathematical foundations have been generally acknowledged to deserve special attention for the best use of this class of numerical methods. In this course the essential aspects of the mathematical theory of finite element methods will be addressed, more particularly in the framework of boundary value problems for ordinary differential equations. For this propose some useful topics of Functional Analysis will be recalled, and a number of classical function spaces will be handled. A first goal of the course is motivating the setting of boundary value problems for differential equations in variational form, followed by a finite element dis- cretization. The sequel is aimed at supplying the basic convergence results for finite element methods in one space dimension. A final objective is understan- ding in main lines, how the material previously presented in the course may be extended to problems posed in two or three space dimensions.

2 Contents

1

Motivation

As it is well-known, the finite element method is employed in most cases, to solve problems of Physics and of Continuum Mechanics governed by boundary value differential equations. In order to illustrate how important it is to know the foundations of this method from the mathematical point of view, we shall begin this course with the study of a problem of the above mentioned type. Although this is rather simple, it will allow us to introduce some concepts of much wider scope. For instance, by just considering such model problem, we will attempt to show the relationship between different possible formulations under which one may write the differential equation, namely, this form itself, the variational form and the underlying minimization problem in some cases. Moreover we will endeavour to give insight on the criteria that lead to the best choice of the type of finite element method, in order to face a given situation encountered in practice. Let us then consider the elongational deformations of a string Ω of lengthL, having one of its ends fixed, under the action of a distribution of forces acting along its length, represented by a functionf(x);0·x·L. We denote byu(x) the resulting length variation of the string at a pointx;or equivalently, the dis- placement undergone in the longitudinal direction of Ω, at this point.

4 1. Motivation

The physical problem just described can be set in the form of a boundary value ordinary differential equation for the unknown functionu(x);known as the

Sturm-Liouville problem

1, namely,8>><

>:¡(pu0)0+qu=fin (0;L) u(0) = 0 (fixed left end) u

0(L) = 0 (free right end)(P

1)

In (P1)u0denotesdu

dx andpandqare functions that represent physical charac- teristics of the material which the string is made of. We shall assume thatpand qsatisfy respectively for certain constants®;¯;AandB: A¸p(x)¸® >0 andB¸q(x)¸¯¸0;8x2[0;L] Moreover temporarily we shall further assume thatpis continuous in (0;L) and also differentiable there, except eventually, at a certain number of points, and thatp0is also bounded, that is, there existsC >0 such thatjp0(x)j ·Cat every pointx2[0;L] wherep0(x) is well defined. Now without caring about regularity assumptions, neither on the datap;qand f;nor on the solutionuof (P1);we shall rewrite this problem in variational form. We only assume that all the operations performed for this purpose are feasible and well defined, leaving for later the discussion about the conditions under which it is really so. Let us then multiply both sides of the differential equation with a "test func- tion"v, and then integrate over the interval (0;L):Using integration by parts we obtain:

¡pu0vjL

0+Z L 0 pu0v0dx+Z L 0 quvdx=Z L 0 fvdx 1

Classically theSturm-Liouville problemstems from the partial differential equation that describes the free

or the forced vibrations of a string. Ifu(x;t) represents the transversal displacement of a particle of the string Ω

located at a pointxat timet, this equation (with suitable initial a boundary conditions) writes½utt¡(rux)x=F

in Ω£]0;1),Frepresenting given forces and½andrbeing physical characteristics of the string depending only on

x:For instance, ifFis periodic in time, say,F(x;t) =eiwtf(x), then one may legitimately setu(x;t) =eiwtu(x),

thereby obtaining the differential equation (P1) foru(x) withp=randq=¡w2½. However in this caseqis

negative, which makes this problem foru(x) fundamentally different from the one to be considered here.

1. Motivation 5

for every (test) functionv. The expression "test function" means thatvis sup- posed to behave somehow like the solutionu, in the sense thatuitself could be one of such functions. We shall simply say thatvsweeps a function spaceV, some properties of which being specified below in order to simplify things. First of all we note that sinceu0(L) = 0, if we require thatvvanish at the origin likeuitself, then our variational problem writes Z L 0 pu0v0dx+Z L 0 quvdx=Z L 0 fvdx;8v2V(P2) whereubelongs also toV. Since all the integrals above must carry a meaning, as we will see later on, it is necessary to require thatftogether with every function inVand its first order derivative are square integrable in the sense of Lebesgue in the interval (0;L). This implies that we are actually definingVto be:

V=fv±v;v02 L2(0;L);v(0) = 0g

whereL2(0;L) =ff/f2has a finite (Lebesgue) integral in (0;L)g

Remark 1.1

As proven later, everyvsuch thatv02 L2(0;L)is continuous. Therefore there is no contradiction in requiring thatv(0) = 0: Now what have we gained by writing the Sturm-Liouville problem (P

1) in the

variational form (P 2)? First of all we note that the highest derivative order that appears in the latter is one, while in the former it is two. Thus for instance, if we choose to approximate uby a function belonging to the class of piecewise polynomials like the finite element method does, it is readily seen that continuous functions may be used for (P

2) whereas continuously differentiable ones would be required for (P1).

Another advantage of formulation (P

2) over (P1) is the fact that the boundary

conditionu0(L) = 0 may be disregarded in the former, for it is implicitly satisfied. Indeed if we still assume that all the operations performed below are legitimate, we have from (P 2). pu 0vjL

0¡Z

L 0 (pu0)0vdx+Z L 0 quvdx=Z L 0 fvdx(1.1)

6 1. Motivation

for everyv2V. Let us choosev2Vsuch thatv(L) = 0, but otherwise arbitrary.

In so doing we get

Z L 0 [¡(pu0)0+qu¡f]vdx;8v2Vsuch thatv(L) = 0 Here the reader should admit that the class os such functionsvis wide enough for the above relation to imply that

¡(pu0)0+qu=falmost everywhere in (0;L)

From this result (1.1) simply becomes

p(L)u0(L)v(L)¡p(0)u0(0)v(0) = 0;8v2V Sincev(0) = 0;choosing nowv2Vsuch thatv(L) = 1 we immediately infer that u

0(L) = 0 as required, aspis strictly positive by assumption.

Another clear advantage of (P

2) over (P1) is related to the regularity of the

datump. Even assuming that the material of the string is homogeneous, the functionpvaries with its cross section. Eventually the latter could change of shape abruptly in such a way thatpis a discontinuous function. While on the one hand this does not bring about any difficulty as far as problem (P

2) is concerned,

much care is needed in handling the term (pu0)0of equation (P1) in this case. Finally a more tricky but fundamental point: formulation (P

2) is more general

than equation (P

1) from the physical point of view. In order to clarify this as-

sertion let us consider a distribution of forcesf"having a resultantP, uniformly acting on a very small length"around the abscissax=L 2 :In current applications such system of forces is represented byP±L 2 , namely, a single force of modulus equal toPapplied at the point given byx=L 2 .2Of course if we stick to the first definition off", both (P1) and (P2) carry a meaning. Actually denoting byu"the 2 In old days many scientists used to call this distribution of forces the "Dirac function"±L 2 multiplied byP, where±L 2 would be such that±L 2 (x) = 0;8x2[0;L]; x6=L 2 ; ±L 2 (L 2 ) = +1and "Z L 0 L 2 dx" = 1.

1. Motivation 7

corresponding solution, the latter writes: Z L 0 pu0"v0dx+Z L 0 qu "vdx=Z L 2 2 L 2 2 P vdx;8v2V Now sincevis a continuous function, the Mean Value Theorem for integrals implies that the right hand side of the above equation tends toPv(L 2 ) as"goes to zero. Otherwise stated, the variational problem (P

2) is perfectly well defined

in the case of the single force applied to a particle located at pointx=L 2 , as Z L 0 pu0v0dx+Z L 0 quvdx=Pv(L 2 );8v2V; uitself inV

On the other hand the equation (P

1) in this case could only be based on an

equality of the type¡(pu0)0+qu=P±L 2 , whose exact meaning lies beyond the scope of ordinary functions. In other words this form is not suitable for being exploited numerically. Since forces applied pointwise may act everywhere and may also be combined, we conclude that there are at least infinitely many system of forces that are admissible for the model (P

2) of the elongational deformation of a string, although

they do not correspond to a differential equation of the form (P

1), at least as far

as functions defined in accordance with the classical concept are concerned. Summarizing we have exhibited at least four advantages of formulation (P 2) over (P

1), in spite of the fact that a rather simple linear ordinary differential

equation was used as a model for this comparative analysis. Let us conclude this chapter with a few considerations about the relationship between smoothness of the data and the choice of the spaceVin which the solution is to be searched for. To begin with we assume again thatpis continuous, and differentiable except eventually at a certain number of points of the interval (0;L) and thatjp0jis bounded by a constantCwherever it is defined. Our goal is to define the largest possible space where the solutionuof (P1) belongs to, under the above assumptions onpandq, by assuming thatfbelongs

8 1. Motivation

toL2(0;L). Of courseushould be at least a continuous function in [0;L], otherwise u

0would carry no meaning in the term¡(pu0)0of the differential equation that

holds in (0;L), and bothu(0) andu0(L) might not be defined.

This implies thatu2 L2(0;L) and moreover since

Z L 0 (qu)2dx·B2Z L 0 u2dx we havequ2 L2(0;L) as well. This implies in turn that¡(pu0)02 L2(0;L), as the difference betweenfandqu: Actually (pu0)0is also (Lebesgue) integrable in (0;L) due to the fact that wheneverfandgare functions inL2(0;L) thenfgis integrable in (0;L). In- deed, sincef§g2 L2(0;L) andZ L 0 (f§g)2dx¸0, we have§Z L 0 fgdx· 1 2 ZL 0 f2dx+Z L 0 g2dx¸ 3. The primitive of the integrable function (pu0)0must be continuous, and hence, sincepis continuous in [0;L] so must beu0. This implies thatu02 L2(0;L) too, as doesu00according to the following argument: Wheneveru00is defined it must be equal to(pu0)0¡p0u0 p . Since the following bound holds: Z L 0· (pu0)0¡p0u0 p 2 dx·2 2½ ZL 0 [(pu0)0]2dx+B2Z L 0 ju0j2dx¾ the functionu00must be defined as an element ofL2(0;L). As a conclusion, wheneverfis square integrable in (0;L) the solutionutogether with its two first derivativesu0andu00are also functions inL2(0;L), providedp;q andp0satisfy our given properties. This leads to the assertion that the right space in which the solution of (P

1) should be searched for is in this case

S=©v±v;v0;v002 L2(0;L); v(0) = 0 andv0(L) = 0ª 3

Now takef= (pu0)0andg´1.

1. Motivation 9

As a matter of factSis a subspace of the classical Sobolev spaceH2(0;L) defined as follows: H

2(0;L) =©v±v;v0;v002 L2(0;L)ª

The additional conditions onv(0) andv0(L) in the definition ofSare applicable to functions inH2(0;L), since those are necessarily continuously differentiable in [0;L] as are can easily check. Now the above argument leading to such spaceScan be extended to the case wherep0is not defined in [0;L]. The conclusion is still that (pu0)02 L2(0;L), which means in particular thatpu0is a continuous function. However in this caseu0will not be continuous in general, though still a function inL2(0;L). This assertion can be easily verified by the reader as an exercise. Then the natural solution space for problem (P

1) becomes

S=©v±v;v0;(pv0)02 L2(0;L); v(0) = (pv0)(L) = 0ª Notice that (pv0)0cannot be viewed as the sum ofpv00andp0v0for everyv2S, since neitherv00norp0are defined in [0;L] in general. Actually when the string cross section undergoes an abrupt change at some point, the displacement"s derivative must also have an abrupt change at this point, in such a way that the productpu0remains continuous there.

As a final comparison between problems (P

1) and (P2), we recall that in the

case of the latter the space in which we must search for the solution is simply the subspaceVof the Sobolev spaceH1(0;L) consisting of those functionsvfor whichv(0) = 0, whereH1(0;L) =fv/v;v02 L2(0;L)g. Of course whenever the right hand side of the variational equality of (P 2) is expressed in the formZ L 0 fvdxwheref2 L2(0;L), its solutionulies indeed in S. However we gain nothing by imposing thatubelong toSbeforehand, simply because as seen before, some conditions in the definition of this space are just too stringent for practical purposes. Moreover if we do so we rule out solutions to (P

2) that do not really belong toS. This is the case whenever the right side

10 1. Motivation

corresponds to forces that are not functions inL2(0;L) or not even functions at all. In order to illustrate this assertion, assume thatq= 0; p= 1 and the right hand side in the variational equality is given byPv(L 2 ). By a direct calculation we can check that the functionugiven by u(x) =8 >:Px;forx2· 0;L 2 P L 2 forx2µL 2 satisfies8v2V: Z L 0 pu0v0dx+Z L 0 quvdx=Z L 2

0Pv0dx

=Pv(L 2 )¡Pv(0) =Pv(L 2 );8v2V

It is hence a solution of (P

2) but belongs by no means toSsinceu00is not defined.

In the sequel we will give additional arguments and results in favor of the choice ofVas an optimal solution space relying upon strictly rigorous mathematical analyses. 2

Some Concepts of Functional Analysis

2.1 Normed Functions Spaces

Before going into a rigorous mathematical analysis of problem (P

2) and its

relation with equation (P

1), it is useful to recall some basic aspects of Functional

Analysis as applied to normed vector spaces.

First of all letVbe a real vector space with null vector~0V.Vis said to be a normed space if it is equipped with a mappingk¢k:V!Rhaving the following properties: (i). kvk ¸0;8v2Vandkvk= 0 if and only ifv=~0V; (ii). k®vk=j®jkvk;8®2R,8v2V; (iii) kv1+v2k · kv1k+kv2k;8v1;v22V; We will mostly be dealing with function vector space. Let us then define as examples, the following norms for the spaceC0[0;L] of continuous functions in

12 2. Some Concepts of Functional Analysis

the closed interval [0;L]: kvk0;1def= maxx2[0;L]jv(x)j(Maximum norm orL1-norm) kvk0;1def=Z L 0 jv(x)jdx(L1-norm) kvk0;2def=· ZL 0 jv(x)j2dx¸ 1 2 (L2-norm) It is well-known that all the above three expressions fulfill the conditions (i)¡ (ii)¡(iii). Actually only property (i) for the normsk¢k0;1andk¢k0;2cannot be established without fully exploiting the continuity of functions inC0[0;L]. This is because such expressions do not really define norms for spaces containing discontinuous functions such asL2(0;L). Indeed, iff2 L2(0;L) andkfk0;2= 0, thenfis a function that vanishesalmost everywherein (0;L), i.e., at every point in this interval but eventually at points that form a set whose measure is equal to zero - in short, a nullset (for instance, a countable set of points). But since the normk¢k0;2is quite natural and handy for square integrable functions, in order to overcome this difficulty, instead ofL2(0;L) itself one usually works with the quotient space L

2(0;L)def=L2(0;L)=N(0;L);

whereN(0;L) is the space of those functions that vanish almost everywhere in (0;L). This is becauseL2(0;L) is indeed a normed space when equipped with the normk¢k0;2(applied to any function representing a given class belonging to this quotient space), for kfk0;2= 0)f2 N(0;L);or yet [f] = [O] whereOdenotes the null function and [f] the class offin the quotient space. Nevertheless, as most authors do, we shall identifyL2(0;L) andL2(0;L) in this text, by abusively regarding the latter as a space of functions not to be distinguished from eachother, as long as they are identical almost everywhere in (0;L).

2.2 Inner Product Spaces 13

Finally we recall that an applicationj¢j:V!Rthat fulfills (ii) and (iii) and such thatjvj ¸0;8v2Vis called asemi-normofV.

2.2 Inner Product Spaces

We shall be particularly concerned about a class of normed spaces whose norm is defined through a given inner product. We recall that a real vector spaceVis said to be equipped with an inner product (¢j¢), if this is a mapping fromV£V ontoRthat satisfies the following properties: (I). (vjv)¸0;8v2Vand (vjv) = 0 if and only ifv=~0V; (II). (vju) = (ujv);8u;v2V; (III). (®1u1+®2u2jv) =®1(u1jv) +®2(u2jv);8®1;®22Rand8u1;u2;v2V. If can be easily checked that to a given inner product (¢j¢) corresponds a norm defined by: kvkdef= (vjv)1 2quotesdbs_dbs6.pdfusesText_12

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