An Introduction to the Mathematical Foundations of the ...€¦ · element method does, it is readily seen that continuous functions may be used for (P2) whereas continuously diﬀerentiable

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 Spaces Hm(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

1Motivation

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 length L,

having one of its ends fixed, under the action of a distribution of forces acting

along its length, represented by a function f(x), 0 ≤ x ≤ L. We denote by u(x)

the resulting length variation of the string at a point x, 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 function u(x), known as the

Sturm-Liouville problem1, namely,

−(pu′)′ + qu = f in (0, L)

u(0) = 0 (fixed left end)

u′(L) = 0 (free right end)

(P1)

In (P1) u′ denotesdu

dxand p and q are functions that represent physical charac-

teristics of the material which the string is made of. We shall assume that p and

q satisfy respectively for certain constants α, β, A and B :

A ≥ p(x) ≥ α > 0 and B ≥ q(x) ≥ β ≥ 0, ∀x ∈ [0, L]

Moreover temporarily we shall further assume that p is continuous in (0, L)

and also differentiable there, except eventually, at a certain number of points,

and that p′ is also bounded, that is, there exists C > 0 such that |p′(x)| ≤ C at

every point x ∈ [0, L] where p′(x) is well defined.

Now without caring about regularity assumptions, neither on the data p, q and

f, nor on the solution u of (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:

−pu′v|L0 +

∫ L

0

pu′v′dx +

∫ L

0

quvdx =

∫ L

0

fvdx

1Classically the Sturm-Liouville problem stems from the partial differential equation that describes the free

or the forced vibrations of a string. If u(x, t) represents the transversal displacement of a particle of the string Ω

located at a point x at time t, this equation (with suitable initial a boundary conditions) writes ρutt−(rux)x = F

in Ω×]0,∞), F representing given forces and ρ and r being physical characteristics of the string depending only on

x. For instance, if F is periodic in time, say, F (x, t) = eiwtf(x), then one may legitimately set u(x, t) = eiwtu(x),

thereby obtaining the differential equation (P1) for u(x) with p = r and q = −w2ρ. However in this case q is

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

1. Motivation 5

for every (test) function v. The expression ”test function” means that v is sup-

posed to behave somehow like the solution u, in the sense that u itself could be

one of such functions. We shall simply say that v sweeps a function space V , some

properties of which being specified below in order to simplify things.

First of all we note that since u′(L) = 0, if we require that v vanish at the

origin like u itself, then our variational problem writes∫ L

0

pu′v′dx +

∫ L

0

quvdx =

∫ L

0

fvdx, ∀v ∈ V (P2)

where u belongs also to V . Since all the integrals above must carry a meaning, as

we will see later on, it is necessary to require that f together with every function

in V and its first order derivative are square integrable in the sense of Lebesgue

in the interval (0, L). This implies that we are actually defining V to be:

V = v /v, v′ ∈ L2(0, L), v(0) = 0

where L2(0, L) = f /f 2 has a finite (Lebesgue) integral in (0, L)

Remark 1.1 As proven later, every v such that v′ ∈ L2(0, L) is continuous.

Therefore there is no contradiction in requiring that v(0) = 0.

Now what have we gained by writing the Sturm-Liouville problem (P1) in the

variational form (P2)?

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

u by 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

(P2) whereas continuously differentiable ones would be required for (P1).

Another advantage of formulation (P2) over (P1) is the fact that the boundary

condition u′(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 (P2).

pu′v|L0 −∫ L

0

(pu′)′vdx +

∫ L

0

quvdx =

∫ L

0

fvdx (1.1)

6 1. Motivation

for every v ∈ V . Let us choose v ∈ V such that v(L) = 0, but otherwise arbitrary.

In so doing we get

∫ L

0

[−(pu′)′ + qu− f ] vdx, ∀v ∈ V such that v(L) = 0

Here the reader should admit that the class os such functions v is wide enough

for the above relation to imply that

−(pu′)′ + qu = f almost everywhere in (0, L)

From this result (1.1) simply becomes

p(L)u′(L)v(L)− p(0)u′(0)v(0) = 0, ∀v ∈ V

Since v(0) = 0, choosing now v ∈ V such that v(L) = 1 we immediately infer that

u′(L) = 0 as required, as p is strictly positive by assumption.

Another clear advantage of (P2) over (P1) is related to the regularity of the

datum p. Even assuming that the material of the string is homogeneous, the

function p varies with its cross section. Eventually the latter could change of

shape abruptly in such a way that p is a discontinuous function. While on the one

hand this does not bring about any difficulty as far as problem (P2) is concerned,

much care is needed in handling the term (pu′)′ of equation (P1) in this case.

Finally a more tricky but fundamental point: formulation (P2) is more general

than equation (P1) from the physical point of view. In order to clarify this as-

sertion let us consider a distribution of forces fε having a resultant P , uniformly

acting on a very small length ε around the abscissa x = L2. In current applications

such system of forces is represented by Pδ L2, namely, a single force of modulus

equal to P applied at the point given by x = L2.2 Of course if we stick to the first

definition of fε, both (P1) and (P2) carry a meaning. Actually denoting by uε the

2In old days many scientists used to call this distribution of forces the “Dirac function” δ L2

multiplied by P ,

where δ L2

would be such that δ L2

(x) = 0,∀x ∈ [0, L], x 6= L2

, δ L2

(L2) = +∞ and “

Z L

0δ L

2dx” = 1.

1. Motivation 7

corresponding solution, the latter writes:

∫ L

0

pu′εv′dx +

∫ L

0

quεvdx =

∫ L2+ ε

2

L2− ε

2

P

εvdx, ∀v ∈ V

Now since v is a continuous function, the Mean Value Theorem for integrals

implies that the right hand side of the above equation tends to Pv(L2) as ε goes

to zero. Otherwise stated, the variational problem (P2) is perfectly well defined

in the case of the single force applied to a particle located at point x = L2, as

∫ L

0

pu′v′dx +

∫ L

0

quvdx = Pv(L

2), ∀v ∈ V, u itself in V

On the other hand the equation (P1) in this case could only be based on an

equality of the type −(pu′)′ + qu = Pδ L2, 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 (P2) of the elongational deformation of a string, although

they do not correspond to a differential equation of the form (P1), 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 (P2)

over (P1), 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 space V in which the

solution is to be searched for.

To begin with we assume again that p is continuous, and differentiable except

eventually at a certain number of points of the interval (0, L) and that |p′| is

bounded by a constant C wherever it is defined.

Our goal is to define the largest possible space where the solution u of (P1)

belongs to, under the above assumptions on p and q, by assuming that f belongs

8 1. Motivation

to L2(0, L). Of course u should be at least a continuous function in [0, L], otherwise

u′ would carry no meaning in the term −(pu′)′ of the differential equation that

holds in (0, L), and both u(0) and u′(L) might not be defined.

This implies that u ∈ L2(0, L) and moreover since

∫ L

0

(qu)2dx ≤ B2

∫ L

0

u2dx

we have qu ∈ L2(0, L) as well. This implies in turn that −(pu′)′ ∈ L2(0, L), as

the difference between f and qu.

Actually (pu′)′ is also (Lebesgue) integrable in (0, L) due to the fact that

whenever f and g are functions in L2(0, L) then fg is integrable in (0, L). In-

deed, since f ± g ∈ L2(0, L) and

∫ L

0

(f ± g)2dx ≥ 0, we have ±∫ L

0

fgdx ≤1

2

[∫ L

0

f 2dx +

∫ L

0

g2dx

]3.

The primitive of the integrable function (pu′)′ must be continuous, and hence,

since p is continuous in [0, L] so must be u′. This implies that u′ ∈ L2(0, L) too,

as does u′′ according to the following argument:

Whenever u′′ is defined it must be equal to(pu′)′ − p′u′

p. Since the following

bound holds:

∫ L

0

[(pu′)′ − p′u′

p

]2

dx ≤ 2

α2

∫ L

0

[(pu′)′]2 dx + B2

∫ L

0

|u′|2 dx

,

the function u′′ must be defined as an element of L2(0, L).

As a conclusion, whenever f is square integrable in (0, L) the solution u together

with its two first derivatives u′ and u′′ are also functions in L2(0, L), provided p, q

and p′ satisfy our given properties. This leads to the assertion that the right space

in which the solution of (P1) should be searched for is in this case

S =v

/v, v′, v′′ ∈ L2(0, L), v(0) = 0 and v′(L) = 0

3Now take f = (pu′)′ and g ≡ 1.

1. Motivation 9

As a matter of fact S is a subspace of the classical Sobolev space H2(0, L)

defined as follows:

H2(0, L) =v

/v, v′, v′′ ∈ L2(0, L)

The additional conditions on v(0) and v′(L) in the definition of S are applicable

to functions in H2(0, L), since those are necessarily continuously differentiable in

[0, L] as are can easily check.

Now the above argument leading to such space S can be extended to the case

where p′ is not defined in [0, L]. The conclusion is still that (pu′)′ ∈ L2(0, L), which

means in particular that pu′ is a continuous function. However in this case u′ will

not be continuous in general, though still a function in L2(0, L). This assertion

can be easily verified by the reader as an exercise. Then the natural solution space

for problem (P1) becomes

S =v

/v, v′, (pv′)′ ∈ L2(0, L), v(0) = (pv′)(L) = 0

Notice that (pv′)′ cannot be viewed as the sum of pv′′ and p′v′ for every v ∈ S,

since neither v′′ nor p′ are 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 product pu′ remains continuous there.

As a final comparison between problems (P1) and (P2), we recall that in the

case of the latter the space in which we must search for the solution is simply

the subspace V of the Sobolev space H1(0, L) consisting of those functions v for

which v(0) = 0, where H1(0, L) = v /v, v′ ∈ L2(0, L).Of course whenever the right hand side of the variational equality of (P2) is

expressed in the form

∫ L

0

fvdx where f ∈ L2(0, L), its solution u lies indeed in

S. However we gain nothing by imposing that u belong to S beforehand, 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 (P2) that do not really belong to S. This is the case whenever the right side

10 1. Motivation

corresponds to forces that are not functions in L2(0, L) or not even functions at

all. In order to illustrate this assertion, assume that q = 0, p = 1 and the right

hand side in the variational equality is given by Pv(L2). By a direct calculation

we can check that the function u given by

u(x) =

Px, for x ∈[0,

L

2

]

PL

2for x ∈

(L

2, L

)

satisfies ∀v ∈ V :

∫ L

0

pu′v′dx +

∫ L

0

quvdx =

∫ L2

0

Pv′dx

= Pv(L2)− Pv(0)

= Pv(L2),∀v ∈ V

It is hence a solution of (P2) but belongs by no means to S since u′′ is not defined.

In the sequel we will give additional arguments and results in favor of the choice

of V as an optimal solution space relying upon strictly rigorous mathematical

analyses.

2Some Concepts of Functional Analysis

2.1 Normed Functions Spaces

Before going into a rigorous mathematical analysis of problem (P2) and its

relation with equation (P1), it is useful to recall some basic aspects of Functional

Analysis as applied to normed vector spaces.

First of all let V be a real vector space with null vector ~0V . V is said to be a

normed space if it is equipped with a mapping ‖·‖ : V → R having the following

properties:

(i). ‖v‖ ≥ 0, ∀v ∈ V and ‖v‖ = 0 if and only if v = ~0V ;

(ii). ‖αv‖ = |α| ‖v‖ , ∀α ∈ R,∀v ∈ V ;

(iii) ‖v1 + v2‖ ≤ ‖v1‖+ ‖v2‖ , ∀v1, v2 ∈ V ;

We will mostly be dealing with function vector space. Let us then define as

examples, the following norms for the space C0[0, L] of continuous functions in

12 2. Some Concepts of Functional Analysis

the closed interval [0, L]:

‖v‖0,∞def= max

x∈[0,L]|v(x)| (Maximum norm or L∞-norm)

‖v‖0,1

def=

∫ L

0

|v(x)| dx (L1-norm)

‖v‖0,2

def=

[∫ L

0

|v(x)|2 dx

] 12

(L2-norm)

It is well-known that all the above three expressions fulfill the conditions (i)−(ii) − (iii). Actually only property (i) for the norms ‖·‖0,1 and ‖·‖0,2 cannot

be established without fully exploiting the continuity of functions in C0[0, L].

This is because such expressions do not really define norms for spaces containing

discontinuous functions such as L2(0, L). Indeed, if f ∈ L2(0, L) and ‖f‖0,2 = 0,

then f is a function that vanishes almost everywhere in (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

norm ‖·‖0,2 is quite natural and handy for square integrable functions, in order

to overcome this difficulty, instead of L2(0, L) itself one usually works with the

quotient space

L2(0, L)def= L2(0, L)/N (0, L),

where N (0, L) is the space of those functions that vanish almost everywhere in

(0, L). This is because L2(0, L) is indeed a normed space when equipped with the

norm ‖·‖0,2 (applied to any function representing a given class belonging to this

quotient space), for

‖f‖0,2 = 0 ⇒ f ∈ N (0, L), or yet [f ] = [O]

where O denotes the null function and [f ] the class of f in the quotient space.

Nevertheless, as most authors do, we shall identify L2(0, L) and L2(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 application |·| : V → R that fulfills (ii) and (iii) and

such that |v| ≥ 0,∀v ∈ V is called a semi-norm of V .

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 space V is

said to be equipped with an inner product (·|·), if this is a mapping from V × V

onto R that satisfies the following properties:

(I). (v|v) ≥ 0, ∀v ∈ V and (v|v) = 0 if and only if v = ~0V ;

(II). (v|u) = (u|v), ∀u, v ∈ V ;

(III). (α1u1 + α2u2|v) = α1(u1|v) + α2(u2|v), ∀α1, α2 ∈ R and ∀u1, u2, v ∈ V .

If can be easily checked that to a given inner product (·|·) corresponds a norm

defined by:

‖v‖ def= (v|v)

12 , ∀v ∈ V

For example the norm ‖·‖0,2 of C0[0, L] is associated with the following inner

product:

(u|v)0def=

∫ L

0

u(x)v(x)dx

The fact that the above expression defines indeed an inner product can be easily

checked and is left as an exercise.

Notice that the expression (·|·)0 is not an inner product for L2(0, L) since pro-

perty (i) does not hold in this case. However like in the case of the norm ‖·‖0,1

and ‖·‖0,2, we may replace L2(0, L) with L2(0, L) over which (·|·)0 becomes indeed

an inner product (provided we regard this space as a function space in the sense

already stated).

In this respect we recall that whenever two functions f and g belong to L2(0, L)

then their product belongs to the space L1(0, L) of (Lebesgue) integrable functions


in (0, L). This result could actually have been established by virtue of the Cauchy-

Schwartz inequality applying to any inner product (·|·) defined over a real vector

space V , namely

|(u|v)| ≤ ‖u‖ ‖v‖ ,∀u, v ∈ V, where ‖v‖ def= (v|v)

12

Indeed, owing to properties (i),(ii) and (iii) we have ∀u, v ∈ V and ∀α ∈ R:

(u|u) + 2α(u|v) + α2(v|v) ≥ 0

Since the quadratic function of α on the left hand side of the above inequality

can only be non-negative if

|2(u|v)|2 − 4(u|u)(v|v) ≤ 0,

the result follows.

Now if suffices to take V = L2(0, L) and inner product (·|·)0 to derive

±∫ L

0

f(x)g(x)dx ≤[∫ L

0

f 2(x)dx

] 12[∫ L

0

g2(x)dx

] 12

,∀f, g ∈ L2(0, L).

Notice that not every norm is associated with a given inner product. In fact

this is true if and only if the norm of V , say ‖·‖, obeys the Parallelogram Law,

namely

‖u + v‖2 + ‖u− v‖2 = 2(‖u‖2 + ‖v‖2) ,∀u, v ∈ V.

In this case the associated inner product is given by

(u|v) =1

4

(‖u + v‖2 − ‖u− v‖2)

As the reader may check, the Parallelogram’s Law for a given inner product can

be established in a straightforward manner. On the other hand, proving that the

above expression defines an inner product whenever the given norm obeys this

law is otherwise complex.

Let us just stress here that to some very useful norms correspond no inner

product, such as the maximum norm. Indeed let us take

u(x) =x

Land v(x) =

(L− x)

L,


both belonging to C0[0, L].

We have ‖u‖0,∞ = ‖v‖0,∞ = 1. Moreover ‖u + v‖0,∞ = 1 and ‖u− v‖0,∞ = 1.

Therefore the Parallelogram Law is violated.

As we should also point out, some norms or inner product are not well suited

to certain spaces, as the corresponding mapping may be unbounded or undefined.

For instance the expression (f |g)0 is not defined for all the functions f and g in

L1(0, L) (take e.g. f(x) = g(x) = x−12 ) and the maximum norm cannot be a norm

for L2(0, L) (take for instance f(x) = x−14 )

Let us now give an application related to the Sturm-Liouville problem:

We saw that a convenient test function space for problem (P2) is the space

V =v

/v ∈ H1(0, L), v(0) = 0

,

where H1(0, L) is the Sobolev space of those functions in L2(0, L) whose first

order derivative also belongs to L2(0, L).

A natural inner product for H1(0, L) is the following one:

(u|v)1def=

∫ L

0

[u(x)v(x) + u′(x)v′(x)] dx

as one can easily verify, with corresponding norm:

‖v‖1,2 = (v|v)121

Notice that the shorter expression well defined for every u, v ∈ H1(0, L), namely

((u|v))1 =

∫ L

0

u′(x)v′(x)dx

corresponds to no norm. Indeed, ((v|v))1 = 0 implies that v is an arbitrary con-

stant in [0, L], which means that actually the expression

|v|1,2

def=

[∫ L

0

[v′(x)]2dx

] 12

defines just a semi-norm over H1(0, L). However over the space of test functions

V , the expression ((·|·))1 defines indeed an inner product and hence |·|1,2 is a norm

of V .

Actually we can even establish the following result:


Proposition 2.1 There exist two strictly positive constants C1 and C2 such that

C1 ‖v‖1,2 ≤ |v|1,2 ≤ C2 ‖v‖1,2 , ∀v ∈ V

The inequality on the right is trivially satisfied with C2 = 1, whereas the one

on the left is a consequence of the following lemma:

Lemma 2.2 (The Friedrichs-Poincare inequality) There exists a constant

Cp such that

‖v‖0,2 ≤ Cp |v|1,2 ,∀v ∈ V

Proof. ∀x ∈ [0, L] we have:

∫ x

0

[v2(y)

]′dy = v2(x),

which implies that

v2(x) = 2

∫ x

0

v(y)v′(y)dy, ∀x ∈ [0, L],

or yet

v2(x) ≤ 2

∫ x

0

|v(y)| |v′(y)| dy, ∀x ∈ [0, L].

Thus ∀x ∈ [0, L]

v2(x) ≤ 2

∫ L

0

|v(y)| |v′(y)| dy ≤ 2 ‖v‖0,2 |v|1,2 ,

from which we obtain

‖v‖20,2 =

∫ L

0

v2(x)dx ≤ 2

∫ L

0

‖v‖0,2 |v|1,2 dx.

This proves the lemma with Cp = 2L. q.e.d. ¥The existence of the constant C1 in Proposition 2.1 directly follows from the

Friedrichs-Poincare inequality with,

C1 = (1 + C2p)−

12 .


Remark 2.3 Two norms ‖·‖a and ‖·‖b are said to be equivalent if there exist two

strictly positive constants C1 and C2 such that for every v in the vector space V

we have

C1 ‖v‖a ≤ ‖v‖b ≤ C2 ‖v‖a

Proposition 2.1 thus means that both norms ‖·‖1,2 and |·|1,2 are equivalent over

the test function space V for problem (P2). ¥

An inner product allows for the extension of the usual geometric concept of

orthogonality to real vector spaces.

Definition 2.4 Let V be a real vector space equipped with inner product (·|·).Two vectors u and v belonging to V are said to be orthogonal if (u|v) = 0. In this

case we write u⊥v.

Definition 2.5 Let S be a subset of a real vector space V equipped with an inner

product (·|·). The orthogonal of S denoted by S⊥ is the set of elements in V which

are orthogonal to every element of S.

Note that the only element in V which is orthogonal to every element of this

space is ~0V . Indeed if this is the case of u ∈ V we have

(u|v) = 0, ∀v ∈ V.

Thus taking v = u we readily conclude that u = ~0V . Therefore ~0V ∈ S⊥ whatever

the set S. Actually S⊥ is always a subspace of V .

Lemma 2.6 (Pythagorean Theorem) If u⊥v then

‖u + v‖2 = ‖u‖2 + ‖v‖2

Proof. Straightforward using the properties of the norm ‖·‖ = ( · |·) 12 . ¥

When the subset S consists of just an element v0 (S = v0) we denote S⊥ by

v⊥0 (rather than v0⊥). Notice that ~0⊥V is V itself and that V ⊥ is ~0V .

Proposition 2.7 Let v0 be a given element of V , v0 6= ~0V . Then ∀v ∈ V there

exists a unique splitting of the form v = αv0 + u where u ∈ v⊥0 , α ∈ R.


Proof. Take α =(v|v0)

‖v0‖2 and u = v − αv0. Uniqueness of this decomposition can

be established as an exercise. ¥

2.3 Closed Sets in Normed Spaces

Let S : N → V be a given sequence (v1, v2, ..., vn, ...) of elements of V also

denoted by vnn or simply vn.

Definition 2.8 A sequence vn contained in a normed real space V is said to

converge to an element v ∈ V if ∀ε > 0 there exists an integer N(ε) ∈ N such

that ‖vn − v‖ < ε ∀n > N(ε). In this case we write vn → v in V .

Clearly the above concept implies that limn→∞

‖vn − v‖ = 0 and due to (iii) we

also have limn→∞

‖vn‖ = ‖v‖ if vn → v, as one can easily check.

Definition 2.9 A subset S of a normed real vector space V is said to be a closed

set if every sequence vn ⊂ S that converges to an element v ∈ V is such that

v ∈ S.

Clearly enough V itself is a closed set, as is the empty set ∅ by definition. A

subspace S of V that is a closed set is said to be a closed subspace of V .

Proposition 2.10 ∀S ⊂ V, S⊥ is a closed subspace of V .

Proof. Let vn ⊂ S⊥ be a sequence converging to v0 ∈ V . We have (vn|v) = 0,

∀v ∈ S and hence (v0 − vn|v) = (v0|v), ∀v ∈ S. Owing to the Cauchy-Schwartz

inequality we then have:

|(v0|v)| ≤ ‖vn − v‖ ‖v‖ ,∀v ∈ S, and ∀n ∈ NSince by assumption ‖vn − v0‖ → 0 we must necessarily have (v0|v) = 0, ∀v ∈ V .

q.e.d. ¥

Definition 2.11 Let S be a subset of a normed real vector space V . The closure

of S denoted by S is the set of all the elements in V that are limits of a sequence

of elements contained in S.

2.3 Closed Sets in Normed Spaces 19

Clearly S ⊂ S and if S is a closed set S = S. However in general S 6= S and

the question is how to characterize S in practical cases. Let us give an example

linked to the Sturm-Liouville problem. We consider the space V of test functions

for problem (P2) and the solution space S for (P1), where

S =v

/v ∈ V, v′′ ∈ L2(0, L), v′(L) = 0

We equip V with the inner product (·|·)1 and we would like to assert the S is V

itself. First of all we note that S ⊂ V, since any converging sequence of S in the

norm of V must converge to an element of V . Next let v be an arbitrary element

of V . We will construct a sequence vn ⊂ S that converges to v, in the following

way.

First for a given m ∈ N we define a function by scaling of v as follows; set

ε = L2m

and

vm(x) =

v[x + ε

(x− L

2

)]for Lε/2

1+ε≤ x ≤ L(1+ε/2)

1+ε

vm

[L(1+ε/2)

1+ε

]= v(L) for L(1+ε/2)

1+ε< x ≤ L

0 for 0 ≤ x < Lε/21+ε

Now we extend vm to the whole R by zero for x < 0 and by vm(L) for x > L. We

still denote this extension by vm.

Notice that vm ∈ V ∀m, but not necessarily to vm ∈ S. Nevertheless we can

“modify” vm by using an auxiliary sequence of functions ρl ⊂ S having a small

support, say, of length O(1l), such as:

ρl(x) =

l(1− 2lx)(lx + 1)2 for − 1l≤ x ≤ 0

0 |x| > 1l

l(1 + 2lx)(lx− 1)2 for 0 ≤ x ≤ 1l


thereby defining for a given m and ∀l such that 1l< Lε/2

1+ε;

vlm(x) =

∫ +∞

−∞vm(y)ρl(y − x)dy,

i.e., by convolution.

Noticing that

∫ +∞

−∞ρl(x)dx = 1,∀l and that vm is continuous, it is not difficult

to establish that ∀x ∈ (0, L), vlm(x) → vm(x) as l goes to infinity. For this it

suffices to figure out that, by the Mean Value Theorem for integrals we have

vlm(x) =

∫ 1l

− 1l

vm(x + z)ρl(z)dz = vm(x +θ

l)

∫ 1l

− 1l

ρl(z)dz,

for a given θ, |θ| < 1. Then the continuity of vm ensures the rest.

2.3 Closed Sets in Normed Spaces 21

Moreover we have ∀x ∈ [0, L]:

∣∣vlm(x)

∣∣2 =

∣∣∣∣∣∫ 1

l

− 1l

vm(x + z)ρl(z)dz

∣∣∣∣∣

2

≤∫ 1

l

− 1l

ρ2l (z)dz

∫ 1l

− 1l

v2m(x + z)dz

≤ 2l

∫ 1l

− 1l

v2m(x + z)dz

= 2l

∫ x+ 1l

x− 1l

v2m(y)dy

≤ 4 maxx∈[0,L]

|v(x)|2

Hence by the Dominated Convergence Theorem (see e.g. [10],WEIR, Lebesgue

Integration and Measure, ’73)∥∥vl

m − vm

∥∥0,2→ 0 as l goes to infinity.

Notice that the above argument allows us to bound above the L2-norm of vlm

independently of l and m by 2√

L ‖v‖0,∞.

Similarly but surely less simply, one can establish that∥∥∥(vl

m

)′ − v′m∥∥∥

0,2con-

verges to zero as well, but here some more sophisticated tools of integration theory

must be employed. More precisely, from (see next Remark):

(vl

m

)′(x) =

∫ +∞

−∞v′m(y)ρl(y − x)dy,

we infer that vlm ∈ L2(0, L),∀l, and if v′m is continuous in [0, L] the same argu-

ment above applies. It can be adapted with rather simple modifications to the

case where (v′m)2 is Riemann-integrable, but not at all if v′m is just an arbitrary

function of L2(0, L). In this case a further regularization is needed, based on the

approximation of v′ by a Riemann-integrable function arbitrarily close to v′ in

the L2-norm.


Remarks:

1. The reader may check that

(vl

m

)′(x) =

∫ +∞

−∞v′m(y)ρl(y − x)dy,

using integration by parts.

2. The possibility to approximate a function in L2(0, L) by a Riemann in-

tegrable function arbitrarily close to it in the ‖·‖0,2 norm is guaranteed

by the fact that L2(0, L) is the completion of the space of Riemann-

integrable functions in (0, L) with respect to this norm as seen in the

next paragraph. ¥

Now notice that

(vl

m

)′(L) =

∫ +∞

−∞v′m(y)ρl(y − L)dy =

∫ L+ 1l

L− 1l

v′m(y)ρl(y − L)dy = 0,

since v′m(y) = 0, ∀y ≥ L− 1

l> L

(1 + ε/2)

1 + ε. Furthermore we have

(vl

m

)′′(x) =

∫ x+ 1l

x− 1l

vm(y)ρ′′l (y − x)dy,

and by a very similar argument to the one employed for bounding above the

integral of(vl

m

)2in (0, L), we can established that the one of

[(vl

m

)′′]2

is bounded,

though not independently of l. Hence we do have vlm ∈ S, ∀l.

Now the sequencevn ⊂ S converging to v ∈ V will be chosen as follows; first

we note that classical results or Sobolev spaces ensure that the scaled sequence

vm converges to v in the norm ‖·‖1,2 (or yet of H1(0, L)). Hence for a given n

one may choose m such that

‖vm − v‖1,2 ≤1

2n.

2.4 Complete Normed Spaces 23

Then for such vm we choose l such that1

l<

Lε/2

1 + εand

∥∥vlm − vm

∥∥1,2≤ 1

2n. In

so doing, from (iii) we see that

∥∥vlm − v

∥∥1,2≤ 1

n.

Finally setting vn = vlm for the pair of indices m and l chosen in this way, we

note that ∀ε0 > 0 its suffices to take N(ε0) as the smallest integer such that

N(ε0) >1

ε0

to have ‖vn − v‖1,2 < ε0,∀n > N(ε0). This implies that V ⊂ S and

hence V = S as asserted.

In the particular example considered above the normed vector space V is the

closure of one of its subspaces S. This situation is often exploited in practical

applications such as convergence analysis of numerical methods, and gives rise to

Definition 2.12 A subset D of a normed vector space V is said to be dense in

another subset C of V , if D ⊂ C and ∀c ∈ C and ∀ε > 0 there exists an element

d ∈ D such that ‖c− d‖ < ε.

Notice that if C is a closed set then any set D ⊂ C that is dense in C satisfies

D = C, and in any case D = C. In the case of the above example we can see that

S is dense in V from the construction of the sequence vn ⊂ S.

2.4 Complete Normed Spaces

The concept of convergence in normed vector spaces implicitly involves the limit

of sequences. However a converging sequence also fits the following one:

Definition 2.13 A sequence vn of a normed space V is a Cauchy sequence

if ∀ε > 0, ∃N(ε) such that ‖vm − vn‖ < ε, ∀m,m > N(ε).

Indeed if vn ⊂ V converges to v ∈ V then ∀ε

2> 0 there exists N(ε) such that

‖vn − v‖ <ε

2and ‖vm − v‖ <

ε

2, provided both m and n are greater than N(ε).

Therefore ‖vn − vm‖ ≤ ‖vn − v‖ + ‖vm − v‖ < ε, ∀m,n > N(ε), and vn is a

Cauchy sequence.


The question then is: would every Cauchy sequence in a normed vector space V

be a converging sequence in this space? The answer is yes only for certain classes

of spaces called complete spaces.

Let us first give two examples showing that the same vector space can be

complete or not depending on the norm it is equipped with.

Consider the space C0[0, L] of continuous functions in [0, L]. First we equip it

with the norm‖·‖0,2, and we consider the sequence fn ⊂ C0[0, L], where fn is

defined as follows: for n = 1, fn(x) = 1, ∀x ∈ [0, L] and for n > 1,

fn(x) =

0 for 0 ≤ x < L2− L

2n2nL

[x− L

2

(1− 1

n

)]for L

2− L

2n≤ x < L

2

1 for L2≤ x ≤ L

Let now ε > 0 be given. For any m,n ∈ N (with m > n for example) we have:

‖fn − fm‖20,2 =

∫ L2− L

2m

L2− L

2n

(2n

Lx− n + 1

)2

dx +

∫ L2

L2− L

2m

(n−m)2

(2x

L− 1

)2

dx

≤(

L

2n− L

2m

) (1− n

m

)2

+L

2m

(n−m

m

)2

=L

2n

(1− n

m

)2

<L

2n

Hence ∀ε > 0 if N(ε) is the smallest integer satisfying N(ε) > 12ε2 we have

‖fn − fm‖0,2 < ε provided both m and n are greater than N(ε), and fn is a

Cauchy sequence in this normed space. However this sequence clearly does not

converge to any element f belonging to C0[0, L], for if such function exists it must

satisfy:∫ L

2− L

2n

0

∣∣f 2(x)∣∣2 dx +

∫ L2

L2− 1

2n

|fn(x)− f(x)|2 dx +

∫ L

L2

[f(x)− 1]2 dx

→ 0

as n goes to ∞. Since f is necessarily bounded as is fn ∀n, it must satisfy:

∫ L2

0

∣∣f 2(x)∣∣2 dx +

∫ L

L2

[1− f(x)]2 dx = 0


This means that f(x) = 0 for 0 ≤ x < L2

and f(x) = 1 for L2

< x ≤ 1! Therefore

the Cauchy sequence fn has no limit in C0[0, L] equipped with the norm ‖·‖0,2.

As a second example we consider the same C0[0, L] equipped with the norm

‖·‖0,∞. Now we claim that such normed space is complete. In order to prove this,

let fn be an arbitrary Cauchy sequence in this normed space. This means that

∀ε > 0,∃N(ε) > 0 such that maxx∈[0,L]

|fn(x)− fn(x)| < ε, ∀m,n > N(ε), x ∈ [0, L].

In particular we have ∀x ∈ [0, L], |fn(x)− fm(x)| < ε, ∀m,n < N(ε), which

means that fn(x) is a Cauchy sequence of R for every x ∈ [0, L]. We have then

found a limiting function f of fn, since the space of real numbers (equipped

with the absolute value as a norm) is a complete space from a well-known axiom.

Thus for every x ∈ [0, L] , fn(x) must have a limit and we call this limit f(x).

The question then is: Is f a continuous function in [0, L]?

By taking an arbitrary point x0 ∈ [0, L] we first note that keeping n = N(ε)+1

and letting m go to infinity we obtain

|fN+1(x)− f(x)| ≤ ε, ∀x ∈ [0, L].

Now since the function fN+1 is continuous at x0 there exists δ > 0 such that

|x− x0| < δ implies that |fN+1(x)− fN+1(x0)| < ε, x ∈ [0, L]. Hence we have

∀x ∈ [0, L] such that |x− x0| < δ:

|f(x)− f(x0)| ≤ |f(x)− fN+1(x)|+|fN+1(x)− fN+1(x0)|+|fN+1(x0)− f(x0)| < 3ε

Thus ∀ε > 0 there exists δ > 0 such that |f(x)− f(x0)| < ε, provided |x− x0| < δ

and x ∈ [0, L] (by taking ε in the above equal to ε/3).

Finally we must establish that fn converges to f in the norm ‖·‖0,∞. Indeed

we know that

|f(x)− fn(x)| ≤ ε,∀x ∈ [0, L] and ∀n > N(ε),

which implies that

maxx∈[0,L]

|f(x)− fn(x)| ≤ ε, ∀n > N(ε).


Hence ∀ε0 > 0 we take N0(ε0) = N(ε0/2) to conclude that ‖fn − f‖0,∞ < ε0,∀n >

N0(ε0).

Remark 2.14 The reader at this point may check that the sequence fn of the

previous example is not a Cauchy sequence of C0[0, L] equipped with the maximum

norm. ¥

We just emphasized that the same space may be complete or not depending on

the norm being considered. However we recall that whenever two norms ‖·‖a and

‖·‖b are equivalent for the same space, then if it is complete for norm ‖·‖a it will

also be complete for norm ‖·‖b

Definition 2.15 A complete normed space, whose norm is associated with an

inner product is called a Hilbert space.

Definition 2.16 A complete normed space whose norm is not associated with an

inner product is called a Banach space.

We just saw that C0[0, L] equipped with the maximum norm is a Banach space,

for we had established that the Parallelogram’s law is violated for this norm. We

also saw that C0[0, L] equipped with the inner product (· |·)0 is not a Hilbert

space. This is however, the case of L2(0, L), as long as we consider it as a space

of classes of functions (a quotient space), as seen hereafter.

Definition 2.17 The completion of a normed vector space V is the space of all

the vectors that are limits of Cauchy sequences of V .

We may define L2(0, L) as the completion of the space of equivalence classes of

functions which only differ from eachother over a null set (a set having a measure

equal to zero), whose squares are Riemann-integrable functions in (0, L). In so

doing L2(0, L) equipped with the inner product (· |·)0 is a Hilbert space. Similarly

L1(0, L) = L1(0, L)/N (0, L) equipped with the ‖·‖0,1 is a Banach space, as the

completion of the space of (equivalence classes of) functions which are Riemann-

integrable in (0, L).

Next we give a result connected to our model problems (P1) and (P2):


Theorem 2.18 The space V = v /v ∈ H1(0, L), v(0) = 0 equipped with the in-

ner product (· |·)1 is a Hilbert space.

Proof. Let vn ⊂ V be a Cauchy sequence for the norm ‖·‖1,2 . This means that

∀ε > 0, ∃N(ε) such that ∀m, n > N(ε)

‖vn − vm‖0,2 < ε and ‖v′n − v′m‖0,2 < ε.

Therefore both vn and [v′n] are Cauchy sequences of L2(0, L) (where [v′n]

denotes the equivalence class of v′n; notice that [vn] contains only vn itself since

this function must be continuous).

Since L2(0, L) is complete for norm ‖·‖0,2 there exists two functions v and w,

where [v] ∈ L2(0, L) and [w] ∈ L2(0, L) such that:

‖vn − v‖0,2 → 0 and ‖v′n − w‖0,2 → 0 as n →∞.

Now we define a function v by

v(x) =

∫ x

0

w(y)dy.

v does belong to V as one can easily check and moreover:

‖vn − v‖20,2 =

∫ L

0

∣∣∣∣∫ x

0

(v′n − w) (x)dy

∣∣∣∣2

dx

≤∫ L

0

[∫ x

0

|(v′n − w) (x)| dy

]2

dx

≤∫ L

0

[∫ L

0

|(v′n − w) (x)| dy

]2

dx

≤∫ L

0

L ‖v′n − w‖20,2 dx

= L2 ‖v′n − w‖20,2


Hence vn → v in L2(0, L) which means that v = v by the uniqueness of the limit.

Finally, since w = v′ = v′ a. e. in [0, L], v ∈ V , and we do have ‖v − vn‖1,2 → 0

as n →∞ as required q.e.d. ¥

Remark 2.19 It can be established (see e. g. [11] ) that there exits a constant C

such that

‖v‖0,∞ ≤ C ‖v‖1,2 ,∀v ∈ H1(0, L) (2.1)

On the basis of inequality (2.1) one can demonstrate as an exercise too, that

H1(0, L) equipped with the inner product (· |·)1 is a Hilbert space. ¥

We conclude this section by clarifying some links between closed spaces and

complete spaces.

As we know every normed vector space V is closed, since it is viewed as a

universe and by definition every converging sequence in V converges necessarily

to an element of V . However V may be complete or not for the given norm. In

this sense both concepts are essentially different. However, whenever a space V is

complete they are equivalent, as far as subspaces of V are concerned, according

to:

Proposition 2.20 If a normed vector space V is complete, then a subspace W

of V is complete if and only if it is closed.

Proof. Let W be complete. If vn ⊂ W converges, say, to v ∈ V, then it is also a

Cauchy sequence of W . Since W is complete the limit v must belong to W which

is thus closed.

Conversely let W be closed and vn ⊂ W be a Cauchy sequence. This is also a

Cauchy sequence of V and since this space is complete there exists v ∈ V such

that vn → v. However W is closed and hence necessarily v ∈ W which implies

that W is complete. q.e.d. ¥As an additional exercise, the reader may establish that the subspace V of

H1(0, L) equipped with the inner product (· |·)1 is a Hilbert space, as a closed

subspace of a Hilbert space. In order to do this inequality (2.1) should be taken

for granted.

2.5 Orthogonal Projections 29

A final result in this section rather easy to establish is the following:

Proposition 2.21 If a subspace W of a complete normed space V is dense in V ,

then the latter is the completion of W with respect to the given norm. ¥

Notice that, as a consequence of Proposition 2.21, the completion of S with

respect to the norm ‖·‖1,2 is V .

2.5 Orthogonal Projections

The setting of problem (P2) like in many others cases of the kind, is strictly

related to the principle of finding a function u that minimizes a certain functional

E over a given function space V . However in practical situation determining u

itself may be out of reach. Then we are lead to an approximate problem in which

instead of u, an approximation of it, say uW is searched for in suitably chosen

subspace W of V . The usual requirement is that uW minimizes the distance from

u to an arbitrary element w ∈ W measured in a norm defined upon an inner

product, that is, for ‖·‖ = (·| ·) 12 :

‖u− uW‖ ≤ ‖u− w‖ ,∀w ∈ W, where uW ∈ W . (2.2)

This inner product is actually a natural one for the problem being considered,

and we will see later on that in the case of (P2) this is nothing but:

(u| v)a =

∫ L

0

p(x)u′(x)v′(x)dx +

∫ L

0

q(x)u(x)v(x)dx.

Posed in such terms we are actually dealing with the problem of finding the

so-called orthogonal projection uW of u ∈ V onto a subspace W of V , that is,

the vector uW such that u − uW ∈ W⊥, where the orthogonality is defined in

connection with an inner product (·| ·) of V . Indeed, assume that there exists

such uW that satisfies (2.2). Then

(u− uW | u− uW ) ≤ (u− w| u− w) , ∀w ∈ W


and in particular for vectors of the form w = uW + εv, v ∈ W and ε ∈ R. Thus

0 ≤ −2ε (u− uW | v) + ε2 (v| v) ,∀v ∈ W and ∀ε ∈ R

or yet

2ε (u− uW | v) ≤ ε2 (v| v) ,∀v ∈ W and ∀ε ∈ R.

Since this implies that |(u− uW | v)| ≤ |ε| (v| v) ,∀v ∈ W and ∀ε ∈ R we conclude

that (u− uW | v) = 0,∀v ∈ W , i.e., u− uW ∈ W⊥.

Conversely if u− uW ∈ W⊥ we have:

(u− uW | w − uW ) = 0,∀w ∈ W

or yet

(u− uW | u− uW ) + (u− uW | w − u) = 0,∀w ∈ W .

Hence

‖u− uW‖2 = (u− uW | u− w) ≤ ‖u− uW‖ ‖u− w‖and uW is seen to satisfy (2.2).

Actually if u − uW ∈ W⊥ then uW is unique, for any other element of W , say

uW∼

such that u− uW∼

∈ W⊥ will satisfy

(u− uW∼|uW − uW

∼) = (u− uW |uW − uW

∼) = 0,

which trivially implies that uW∼

= uW .

We are then left the following question: Under which assumptions on V and

W there exists uW satisfying (2.2) for any given u ∈ V ? A partial answer to it is

provided by the following Theorem:

Theorem 2.22 (The Classical Theorem of the Orthogonal Projection) If

V is a Hilbert space and W is a closed subspace of V for every u ∈ V there exists

a unique uW ∈ W that minimizes ‖u− v‖ over W .

Proof. Let δ = infw∈W

‖u− w‖ and wn ⊂ W be a minimizing sequence, that is:

‖u− wn‖ → δ as n →∞.

2.6 Bounded Linear Functionals 31

By the Parallelogram law we have ∀m, n ∈ N:

2(‖u− wn‖2 + ‖u− wm‖2) = 4

∥∥∥∥u− wn + wm

2

∥∥∥∥2

+ ‖wn − wm‖2

Since W is a subspace of V , ‖u− (wn + wm) /2‖2 ≥ δ2 and hence

‖wn − wm‖2 ≤ 2[(‖u− wn‖2 − δ2

)+

(‖u− wm‖2 − δ2)]

.

Now for a given ε > 0 it is possible to choose N(ε) such that ‖u− w2n‖ − δ2 <

ε, ∀n > N(ε). Hence for a given ε0 > 0 there exists N0(ε0), say, N(ε20

4), such

that ‖wn − wm‖ < ε0,∀m,n > N0(ε0). The minimizing sequence wn is hence

a Cauchy sequence of W and since this is a complete subspace of V because it

is closed and V is a Hilbert space, there exists an element uW ∈ W such that

wn → uW . Noticing that necessarily ‖u− wn‖ converges to ‖u− uW‖, we have

δ = ‖u− uW‖ ≤ ‖u− w‖ ,∀w ∈ W as required.

The fact that uW is unique was already established. q.e.d. ¥

2.6 Bounded Linear Functionals

In many applications such as the deformation of a string described by the

Sturm-Liouville problem, the physical data appear in the form of a linear func-

tional F : V → R. where V is a space of functions representing unknown fields,

in which the actual solution must be searched for. In general V is equipped with

a given natural norm, say, ‖·‖, and it is required that F be bounded, according

to the following:

Definition 2.23 A functional F : V → R, where V is a normed vector space is

bounded if

∃C > 0 such that |F (v)| ≤ C ‖v‖ ,∀v ∈ V

Typical examples of bounded functionals may be found precisely in the frame-

work of our model problem (P2), for which V was defined in Chapter 1, and we

considered two types of physical data (loads):


• A density of forces f per unit of length → F (v) =

∫ L

0

fvdx, f ∈ L2(0, L);

• A force acting pointwise (e.g. at x = L2) → F (v) = Pv(L

2), P ∈ R

Notice that if we equip V with the norm ‖·‖1,2, in both cases F is bounded.

Indeed: ∣∣∣∣∫ L

0

fvdx

∣∣∣∣ ≤ ‖f‖0,2 ‖v‖0,2 ≤ ‖f‖0,2 ‖v‖1,2 ⇒ C = ‖f‖0,2 ;

∣∣∣∣Pv

(L

2

)∣∣∣∣ ≤ |P | C ‖v‖1,2 ⇒ C = PC, where C verifies (2.1).

Now we recall that a given functional F applied to elements of a normed

vector space V is continuous at a given v0 ∈ V if ∀ε > 0,∃δ(ε, v0) such that

|F (v)− F (v0)| < ε, ∀v satisfying ‖v − v0‖ < δ. Notice that this definition is

equivalent to the following one:

F is continuous at v0 ∈ V if every sequence vn converging to v0 in V satisfies

F (vn) → F (v0) in R.

A strong result that can be established for linear functionals is the following:

Proposition 2.24 A linear functional F on a normed vector space V is contin-

uous at every v ∈ V if and only if it is bounded.

Proof. First of all we note that since F is linear, if it is continuous at ~0V it will

be continuous at every v ∈ V . Indeed if F is continuous at ~0V then ∀ε > 0, ∃δsuch that |F (v)| < ε if ‖v‖ < δ. Then ∀v0 ∈ V , |F (v)− F (v0)| < ε provided

‖v − v0‖ < δ since F (v)− F (v0) = F (v − v0) by linearity.

Now assuming that F is bounded, ∀v ∈ V and ∀ε > 0 it suffices to take δ =ε

C to

establish the continuity at ~0V .

Conversely, it F is continuous take a given ε > 0 and a corresponding δ > 0 such

that ‖v‖ < δ ⇒ |F (v)| < ε. Now for an arbitrary v′ ∈ V let v = δv′/2 ‖v′‖ if

v′ 6= ~0V and v = ~0V otherwise. Clearly |F (v)| < ε and by the linearity of F we

have:δ

2

|F (v′)|‖v′‖ < ε for v′ 6= ~0V ,


and in any case |F (v′)| ≤ C ‖v′‖ , ∀v′ ∈ V with C = 2εδ. q.e.d. ¥

At this point it seems interesting to show an example of a linear functional which

is not bounded on our space V of test functions for Problem (P2): Take L = 1

and set F (v) = v′(L). Now consider the sequence vn ⊂ V with vn(x) =xn

n. We

have F (vn) = 1, ∀n and |vn|1,2 = (2n− 1)−12 . Thus according to Proposition 2.1,

‖vn‖1,2 tends to zero as n → ∞ and hence this functional is unbounded (this is

not surprising since there are functions in V whose first order derivative are not

bounded in (0, 1), such as v(x) = (1− x)34 ).

In more physical terms, this means that “pinching” loads, or yet “dipoles”,

yielding a right hand side of (P2) of the form Qv′(x0), 0 ≤ x0 ≤ L, (where x0

is the point where the string is “pinched”), Q ∈ R are not admissible to this

problem. On the other hand, as already stated, loads applied pointwise, that is

those of the “Dirac” type, corresponding to functionals of the form Pv(x0) are

bounded in V equipped with the ‖·‖1,2 norm and therefore admissible to problem

(P2).

Definition 2.25 The kernel of a given functional F : V → R is the set

Ker(F ) = v /v ∈ V, F (v) = 0.

Clearly if F is a linear functional then Ker(F ) is a subspace of V called the

nullspace of F . If V is a normed space it is easy to prove that Ker(F ) is a closed

subspace of V .

Now we conclude this Chapter by giving a classical and powerful result related

to bounded linear functionals on Hilbert spaces.

Theorem 2.26 (The Riesz Representation Theorem) Let V be a Hilbert

space for inner product (·| ·) and F : V → R be a bounded linear functional. Then

there exists a unique vF ∈ V such that F (v) = (vF | v) , ∀v ∈ V and moreover

supv∈Vv 6=~0V

F (v)

‖v‖ = ‖vF‖ .


Proof. If F (v) = 0,∀v ∈ V then clearly vF = ~0V fulfills all the required properties.

Otherwise Ker(F ) is not the whole V , and we may select a w ∈ V such that

w /∈Ker(F ).

Since Ker(F ) is closed, from the Theorem of the Orthogonal Projection there

exists w0 ∈ Ker(F ) such that z = w − w0 ∈ Ker(F )⊥ with z 6= ~0V . Hence, noting

that F (z) = F (w) 6= 0 we can define z′ =z

F (z), z′ ∈ Ker(F )⊥.

Let now v be an arbitrary element of V . By linearity we have v−F (v)z′ ∈ Ker(F )

and on the other hand,

(v − F (v)z′| z′) = 0 since z′ ∈ Ker(F )⊥.

Therefore (v| z′) = F (v) (z′| z′), and since z′ 6= ~0V we have

F (v) = (vF | v) , ∀v ∈ V , with vF =z′

‖z′‖2 .

Furthermore we have:

‖vF‖ =(vF | vF )

‖vF‖ ≤ supv∈Vv 6=~0V

(vF | v)

‖v‖ = supv∈Vv 6=~0V

F (v)

‖v‖ ≤ supv∈Vv 6=~0V

‖vF‖ ‖v‖‖v‖ = ‖vF‖ .

Finally if vF ∈ V is such that F (v) = (vF | v) , ∀v ∈ V , then ∀v ∈ V (vF − v′F | v) =

0, and hence vF = v′F . q.e.d. ¥

Remark 2.27 Notice that for a bounded linear functional F with

F (v) ≤ C ‖v‖ ,∀v ∈ V ,

we have

CF = supv∈Vv 6=~0v

F (v)

‖v‖ ≤ C,

and also F (v) ≤ CF ‖v‖ ,∀v ∈ V . Otherwise stated

CF = inf C ∈ R /F (v) ≤ C ‖v‖ ,∀v ∈ V ,


and this infimum is necessarily a minimum given by ‖vF‖. CF is actually a norm

of F in the vector space V ′ consisting of all bounded linear functionals defined on

V (cf. [6],’69), the basic operations on it being defined as follows:

For F1, F2 ∈ V ′, F1 + F2 is given by (F1 + F2)(v) = F1(v) + F2(v),∀v ∈ V ;

For F ∈ V ′ and α ∈ R, αF is given by (αF )(v) = αF (v), ∀v ∈ V ;

The thus normed space V ′ is called the dual space1 of V . The Riesz Representa-

tion Theorem then states that for every Hilbert space V there exists an isometry

between itself and its dual space. ¥

The reader may practice the above concepts by searching for the function vF ∈H1(0, L) such that

(vF | v)1 = F (v),∀v ∈ H1(0, L),

where

F (v) = v(L

2) +

∫ L

0

fvdx, with f(x) =

x for 0 ≤ x < L

2L2

for L2≤ x < L

.

1Some authors prefer to call it the topological dual space of V thereby stressing the fact that it is associated

with a given metrics.


3The Finite Element Method in One Dimension

Space

In this Chapter we shall introduce the classical finite element method to solve

ordinary differential equations with given boundary values, taking the Sturm-

Liouville problem as a model. Since this is a numerical solution procedure we will

also be concerned about the evaluation of the underlying approximation error,

together with the convergence of successively computed solutions to the exact

one as one goes into an ideal limiting process.

Before doing so we will study a general functional framework for boundary

value differential equations among others problems.

3.1 An Abstract Linear Variational Problem

Let V be a vector space equipped with inner product (·| ·) and a : V × V → Rand F : V → R be two mappings satisfying the following assumptions, in which

‖·‖ denotes the norm given by (·| ·) 12 .

38 3. The Finite Element Method in One Dimension Space

Assumption 1

(a). a is bilinear, i.e., linear with respect to each argument;

(b). a is bounded: ∃M > 0 such that a(u, v) ≤ M ‖u‖ ‖v‖ , ∀u, v ∈ V ;

(c). a is symmetric: a(u, v) = a(v, u), ∀u, v ∈ V ;

(d). a is coercive: ∃µ > 0 such that a(v, v) ≥ µ ‖v‖2 , ∀v ∈ V .

Assumption 2

(e). F is linear;

(f). F is bounded.

Now we consider the general linear variational problem, of the form

Find u ∈ V such that

a(u, v) = F (v),∀ ∈ V(3.1)

together with the following “approximate” one, in the sense that its solution

provides an approximation of u: Letting W ⊂ V be a given subspace of V , we

wish to solve Find uW ∈ W such that

a(uW , w) = F (w),∀w ∈ W(3.2)

For problems (3.1) and (3.2) we have the following:

Theorem 3.1 Under the Assumptions 1 and 2 both problem (3.1) and (3.2) have

a unique solution provided V is a Hilbert space equipped with inner product (·| ·)and W is closed. Moreover in this case the following “error” bound applies:

‖u− uW‖ ≤√

M

µinf

w∈W‖u− w‖ . (3.3)

Proof. First we note that bilinear form a defines an inner product on V , according

to Assumption 1. The underlying norm denoted by ‖·‖a = a(·, ·) 12 is equivalent to

the standard one ‖·‖, since

µ ‖v‖2 ≤ ‖v‖2a ≤ M ‖v‖2 ,∀v ∈ V

3.1 An Abstract Linear Variational Problem 39

Therefore V is also complete when equipped with norm ‖·‖a, and hence it is also

a Hilbert space for the inner product given by a(·, ·).Now from the Riesz Representation Theorem we know that there exists a unique

v ∈ V such that F (v) = a(vaF , v),∀v ∈ V . Hence problem (3.1) is equivalent to

Find u ∈ V such that

a(u, v) = a(vaF , v),∀v ∈ V

This problem has clearly a unique solution, namely, u = vaF .

Now we consider problem (3.2). Since W is a closed subspace of Hilbert space

V for the inner product (·| ·), it is also a closed subspace for the inner product

a(·, ·) due to the equivalence of the corresponding norms. Hence W is also a

Hilbert space for inner product a(·, ·) and here again by the Riesz Representation

Theorem restricted to the closed subspace W, problem (3.2) also has a unique

solution, since W is necessarily complete.

Actually noticing that

F (w) = a(u,w) = a(uW , w),∀w ∈ W

we infer that uW is nothing but the orthogonal projection of u onto the closed

subspace W for the inner product a(·, ·). Therefore we have:

a(u− uW , u− uW ) ≤ a(u− w, u− w),∀w ∈ W ,

which yields

µ ‖u− uW‖2 ≤ M ‖u− w‖2 ,∀w ∈ W ,

and finally (3.3). q.e.d. ¥Now we introduce an energy functional related to problem (3.1),

namely E : V → R, with

E(v) =1

2a(v, v)− F (v).

The above definition gives rise to the following result:


Theorem 3.2 If V is a Hilbert space for inner product (·| ·), under Assumptions

1 and 2 there exists a unique u ∈ V that minimizes E over V , that is, the solution

of Find u ∈ V such that

E(u) ≤ E(v), ∀v ∈ V(3.4)

Moreover u is precisely the solution of (3.1).

Proof. Let u be the solution of (3.1). We have

a(u, u− v) = F (u− v), ∀v ∈ V

and from (a).(c),(d) and (e) we derive successively:

a(u, u− v) ≤ F (u− v) +1

2a(u− v, u− v),∀v ∈ V

a(u, u)− a(u, v) ≤ F (u)− F (v) +1

2a(u, u)− a(u, v) +

1

2a(v, v),∀v ∈ V

Hence

E(u) ≤ E(v), ∀v ∈ V .

The existence of u that solves (3.4) having been established all that is left to do is

proving that this is the only solution of this problem. Let then u ∈ V be another

solution of (3.4). We have simultaneously

E(u) ≤ E(u + u

2) and E(u) ≤ E(

u + u

2)

Adding up and developing both inequalities we easily obtain:

1

2a(u, u) +

1

2a(u, u) ≤ 1

4[a(u, u) + a(u, u) + a(u, u) + a(u, u)]

which yields1

4a(u− u, u− u) ≤ 0.

Finally from (d) we derive u = u. q.e.d. ¥

3.1 An Abstract Linear Variational Problem 41

Remarks:

1. We have actually established the equivalence of (3.1) and (3.4) under

Assumptions 1 and 2, without taking into account the fact that V is a

Hilbert space. However this assumption is essential in order to ensure

that the solution to (3.1), and hence to (3.4), does exist.

2. (c) is not essential for the existence and uniqueness of a solution to

(3.1). Actually the case where a is not symmetric is addressed by the

celebrated Lax-Milgram Theorem (cf. [2] ’76). However in this case

problem (3.4) does not make sense, and there is no equivalence between

(3.1) and the minimization of an energy functional in general.

3. If a does not satisfy (c) we can still establish the following error bound

for problem (3.2): ‖u− uW‖ ≤ M

µinf

w∈W‖u− w‖. ¥

Let us now apply the results of theorems (3.1) and (3.2) to the Sturm-Liouville

problem. First we recall that problem (P2) fits into general form (3.1) if we set

a(u, v) =

∫ L

0

pu′v′dx +

∫ L

0

quvdx (3.5)

where the space V is the one defined in Chapter 1. On the other hand, the linear

functional F may take any form for which there exists a constant C such that

F (v) ≤ C ‖v‖1,2 ,∀v ∈ V (3.6)

We recall that V is a closed subspace of H1(0, L) equipped with the ‖·‖1,2 norm,

and is hence a Hilbert space. Note also that the form a given by (3.5) fulfills all

the four conditions specified in Assumption 1, provided p is bounded above and

below by strictly positive constants A and α, and that B ≥ q ≥ 0, ∀x ∈ [0, L] In

particular in this case we have

a(u, v) ≤ M ‖u‖1,2 ‖v‖1,2 ,∀u, v ∈ V


and

a(v, v) ≥ µ ‖v‖21,2 ,∀v ∈ V

with M = max(A,B) and µ = αC21 , where C1 is one of the two constants that

appear in Proposition 2.1 of Chapter 2.

Taking all these facts into account, according to Theorem 3.1, problem (P2)

has a unique solution u ∈ V . Notice that this can eventually belong to one of

the spaces S defined in Chapter 1 or even to a smaller space, depending on the

regularity of the data f, p and q (for instance, if f ∈ H1(0, L), p′ ∈ C0[0, L] and q ∈H1(0, L) we have u ∈ S∩H3(0, L), whereH3(0, L) = v /v, v′, v′′, v′′′ ∈ L2(0, L)).However we gain nothing by searching u in such spaces, which incidentally are

not Hilbert spaces for the natural inner product (·| ·)1. This is because one might

be facing the non existence of a solution in doubtful cases. Moreover in any case

if the solution is more regular than functions belonging “only” to H1(0, L) it will

find itself in the right space naturally.

Let us conclude this section by applying the result of Theorem 3.2: The solution

u ∈ V is the one that minimizes the total energy of the string under the action

of the forces corresponding to functional F that is

E(u) ≤ E(v),∀v ∈ V

with

E(v) =1

2

[∫ L

0

p(v′)2dx +

∫ L

0

qv2dx

]− F (v).

3.2 The Ritz Approximation Method

Now we are ready to go into the main purpose of these lectures by considering

first the simplest possible finite element method in use to solve second order

ordinary differential equations corresponding to a minimization of a (convex)

functional such as (P1), namely, piecewise linear approximation methods. The

main principle stems from the celebrated Ritz method, which is based on the

3.2 The Ritz Approximation Method 43

definition of a finite dimensional subspace of V , say W = VN , where N is the

dimension of VN . This gives rise to the following approximate problem:

Find uN ∈ VN such that

a(uN , v) = F (v),∀v ∈ VN

(3.7)

According to Theorem 3.1, (3.7) has a unique solution, since any finite di-

mensional subspace is closed in any norm one may equip the whole space V (1).

Moreover, if ϕiNi=1 is a basis of VN then there exists a unique set of N constants

c1, c2, . . . , cN such that uN =n∑

j=1

cjϕj.

Since we may take as test function v ∈ VN either one of the functions ϕi,

problem (3.7) yields the following system of equations for the unknowns cj, j =

1, 2, . . . , N :

a

(N∑

j=1

cjϕj,ϕi

)= F (ϕi) for i = 1, 2, . . . , N,

or yet by linearity:

N∑j=1

cja(ϕj,ϕi) = F (ϕi) for i = 1, 2, . . . , N, (3.8)

This is nothing but the linear system of equations

A~c = ~f (3.9)

with ~c = (c1, c2, . . . , cN), where A = aij is an n× n matrix with aij = a(ϕj,ϕi

)

and ~f ∈ RN is given by:

~f = (F (ϕ1), F (ϕ2) . . . , F (ϕN)).

The other way around, one may “multiply” each equation of (3.8) with an

arbitrary scalar vi ∈ R and add them up from i = 1 up to i = N , thereby

1This is an easy-to-prove consequence of the completeness of RN , which is in turn a consequence of the

completeness of R.


obtaining again by linearity:

a

(uN,

N∑i=1

viϕi

)= F (

N∑i=1

viϕi),∀~v = (v1, v2, . . . , vN) ∈ RN ,

which brings us back to problem (3.7) since every element v in VN may be written

in the form v =N∑

i=1

viϕi for suitable scalars vi. Otherwise stated problem (3.7) is

equivalent to system (3.8) or yet to (3.9).

Remark 3.3 We do not need to require the coerciveness of bilinear form a, in

order to apply the principles of Ritz’s method. As a matter of fact, any linear

differential equation of the form D(u) = f , where D is a differential operator and

f is a given function, may be written in an equivalent variational form, that is, a

weak form, as follows:

(D(u)| v) = (f | v) ,∀v ∈ V

for a suitable inner product (·| ·) and a suitable space V .

If the expression (D(u)| v) gives rise to an appropriate bilinear form, say, a(u, v) =

(D(u)| v) , ∀u, v ∈ V , which has the properties (a) and (b), but not necessarily

(c) and (d), and F (v) = (f | v) ,∀v ∈ V has the properties (f), we may still ap-

proximate the corresponding problem of the form (3.1) by (3.7). In so doing we are

actually employing the so-called Galerkin method or yet the Ritz-Galerkin

method. ¥

Let us now turn our attention to the practical solution of (3.7). The equivalent

linear system (3.8) or (3.9) has an invertible matrix, since problem (3.7) is well-

posed. As a matter of fact matrix A is symmetric and positive-definite, and hence

there should be no serious problem to solve it. However this assertion somewhat

masks the fact that the choice of the basis ϕi plays certainly a crucial role, as

far as computational complexity and round-off error control are concerned, among

other practical aspects such as storage requirements.

3.3 Piecewise Linear Finite Element Approximations 45

Just to illustrate the latter point, assume that we are able to construct an

orthogonal basis ϕi with respect to the energy inner product a(·, ·) Then the

solution ~c to (3.9) will appear explicitly, that is: ci =fi

a(ϕi, ϕi), i = 1, 2, . . . , N .

However the construction of such basis is in most cases much too complex to

justify its use. Instead one could be satisfied with the use of an easy-to-handle

basis for which a(ϕj, ϕi) vanishes for a very large number of pairs of indices (i, j).

This is precisely the case of the finite element method, whose basic version is

presented in the following section.

3.3 Piecewise Linear Finite Element Approximations

In its most classical version, the piecewise linear finite element method corre-

sponds to the following choice of a basis in the Ritz method. First we subdivide

the interval [0, L] into N equal intervals (or “elements”) of length h =L

N. Then

we associate with every interval end (or “node”) xi = ih, a function ϕi having

the following properties for i = 0, 1, . . . , N :

1. ϕi restricted to any interval (xj−1, xj) is a polynomial of degree less than or

equal to one, j = 1, 2, . . . , N .

2. ϕi(xj) = δij.

The figure below illustrates the aspect of a generic ϕi:


It is an easy matter to prove that the so-defined functions ϕ0, ϕ1, . . . , ϕN are

linearly independent. Indeed ifN∑

j=0

αjϕj ≡ 0 thenN∑

j=0

αjϕj(xi) = 0,∀i, 0 ≤ i ≤ N ,

and hence αi = 0,∀i. Actually the set ϕiNi=1 forms a basis of the following space:

Vh =

v/

v ∈ C0[0, L], v(0) = 0, v|[xi−1,xi]∈ P1, for i = 1, 2, . . . , N

,

where Pk denotes the space of polynomials (in one variable) of degree less than or

equal to k. Notice that Vh is a subspace of V , for in the case of spaces of piecewise

polynomial functions v, we have v ∈ C0[0, L] ⇔ v ∈ H1(0, L), as one may easily

check.

Remark 3.4 Since ϕi(0) = 0,∀i ≥ 1 and every v ∈ Vh is easily found to be equal

toN∑

i=1

v(xi)ϕi, the set ϕiNi=1 is indeed a basis of Vh. ¥

Now we consider the application of the Ritz-method to the case where VN is the

particular space Vh defined as above. This gives rise to the following approximation

of (P2): Find uh ∈ Vh such that

a(uh, v) = F (v),∀v ∈ Vh

(3.10)

where a is the form given by (3.5) and F is a linear functional satisfying (f) or

yet (3.5).

Clearly problem (3.10) is of the form (3.2) with W = Vh. Hence according to

Theorem 3.1 it has a unique solution, and moreover we have:

‖u− uh‖1,2 ≤√

M

µinf

v∈Vh

‖u− v‖1,2 . (3.11)

We recall that M and µ are given by M = max(A,B) and µ = αC21 where C1 is

defined in Proposition 2.1, Chapter 2, A = ‖p‖0,∞ , B = ‖q‖0,∞.

Before turning our attention to the convergence analysis of the finite element

method, we find it instructive to give an idea of the aspect matrix A and vector


~f take in practical situations. For this purpose we consider, the particular case

where p, q and f are constant. Taking N = 4, for instance, we have:

A =

2ph

+ 2hq3

− ph

+ qh6

0 0

− ph

+ qh6

2ph

+ 2qh3

− ph

+ qh6

0

0 − ph

+ qh6

2ph

+ 2qh3

− ph

+ qh6

0 0 − ph

+ qh6

ph

+ qh3

and

~f =[

fh fh fh fh2

]T

.

The reader may easily find out that for any N,N ≥ 2, we have:

aij = 0 if |i− j| > 1, and aii =2p

h+

2qh

3for 1 ≤ i ≤ N − 1,

aNN =p

h+

qh

3, ai,i+1 = −p

h+

qh

6for 1 ≤ i ≤ N − 1, and

aij = aji,∀i, j, while fi = fh for 1 ≤ i ≤ N − 1 and fN =fh

2.

Notice that A is indeed a sparse matrix (at least for N ≥ 4), more precisely a

tridiagonal matrix.

Now in order to establish the convergence of uh to u in V as h goes to zero, we

address the basic issue in the mathematical analysis of the finite element method,

namely, deriving a possibly optimal upper bound for the term infv∈Vh

‖u− v‖1,2.

Clearly, according to the Theorem of the Orthogonal Projection, this infimum is

actually a minimum attained for v = Πh(u), namely, the orthogonal projection

onto Vh of the exact solution u of (P2), with respect to the inner product (· |·)1 .

However it turns out that the best we can hope for, at least qualitatively in terms

of estimates in the Sobolev norms, is attained if we just take the upper bound

of this minimum given by ‖u− Ih(u)‖1,2, where Ih(u) is the function of Vh that

interpolates the (continuous) solution of (P2) at the nodes xi (of the “mesh”

[xi−1, xi]Ni=1), i = 0, 1, . . . , N .

Let us then derive an approximate estimate for the term ‖u− Ih(u)‖1,2. We

have: ‖u− Ih(u)‖21,2 =

N∑i=1

[‖u− Ih,i(u)‖2

0,2,ki+ ‖u′ − I ′h.i(u)‖2

0,2,ki

], where Ih,i(u)


is the restriction of Ih(u) to the “element” Ki = [xi−1, xi], i = 1, 2, . . . , N , and

‖·‖0,2,Kidenotes the norm of L2(Ki), i.e.,

‖·‖0,2,Ki=

[∫ xi

xi−1

(·)] 1

2

At this point we first assume that the solution u of (P2) belongs to H2(0, L).

In so doing we define ∆i = u − Ih,i(u) and we note that ∆i(xi) = ∆i(xi−1) = 0.

Hence we may legitimately consider Fourier’s sinus series to expand ∆i, that is:

∆i(x) =∞∑

m=1

cim sin

mπ(x− xi−1)

h, ci

m ∈ R,∀m.

Performing the change of variable y = x− xi−1 and setting ∆i(y) = ∆i(x), we

have: ∫ xi

xi−1

∆2i dx =

∫ h

0

∆i(y)dy =h

2

∞∑m=1

(cim

)2,

since according to well-known results,

∫ h

0

cosmπy

hcos

nπy

hdy =

∫ h

0

sinmπy

hsin

nπy

hdy =

h

2δmn, ∀m,n ∈ Z.

For this reason we also have:

∆′i(x) = ∆′

i(y) =∞∑

m=1

cim

mπ

hcos

mπy

h⇒

∫ xi

xi−1

(∆′i)

2dx =

h

2

∞∑m=1

(cim

)2 m2π2

h2

and

∆′′i (x) = ∆′′

i (y) = −∞∑

m=1

cim

m2π2

h2sin

mπy

h⇒

∫ xi

xi−1

(∆′′i )

2dx =

h

2

∞∑m=1

(cim

)2 m4π4

h4.

Notice that since uh is linear in [xi−1, xi] we have ∆′′i (x) = u′′(x),∀x ∈ Ki, and

moreover from the fact that m ≥ 1 we have

‖∆′i‖0,2,Ki

≤∞∑

m=1

(cim

)2 m4π2

h2=

h2

π2‖u′′‖2

0,2,ki, ∀i, i = 1, 2, . . . , N.


Therefore summing up from i = 1 up to i = N we obtain

‖u′ − I ′h(u)‖20,2 ≤

h2

π2‖u′′‖2

0,2

A similar calculation leads to

‖u− Ih(u)‖20,2 ≤

h4

π4‖u′′‖2

0,2

which yields

‖u− Ih(u)‖1,2 ≤h

π

[1 +

h2

π2

] 12

|u|2,2 , (3.12)

where |·|2,2 is the seminorm of H2(0, L) given by |v|2,2 = ‖v′′‖0,2.

This estimate immediately allows us to state that both error terms ‖u− uh‖0,2

and ‖u′ − u′h‖0,2 are at least an O(h) term. However as far as the former is con-

cerned, this is unsatisfactory, since one should expect a better estimate for this

term than the one that holds for first order derivatives. That is why we further

refine this estimate of ‖u− uh‖0,2 by using a duality argument known as Nitsche’s

trick. This can be applied under the assumption that p′ is bounded, for we know

that in this case the solution v of the Sturm-Liouville problem of type (P1) be-

longs to H2(0, L) for any right hand side g in L2(0, L). Furthermore it can also

be established quite easily, that there exists a constant C such that

|v|2,2 ≤ C ‖g‖0,2 ,∀g ∈ L2(0, L) (3.13)

(where C may be taken equal to√

3[1 +

(CC2

p + BC4p

)/α2

] 12 /α, Cp being the

constant of the Friedrichs-Poincare inequality and C being the upper bound of

|p′|).Now we first note that ‖u− uh‖0,2 = sup

g∈L2(0,L)g 6=0

(u− uh |g)0

‖g‖0,2

.

Hence if for a given g ∈ L2(0, L), v ∈ H2(0, L) satisfies

−(pv′)′ + qv = g, v(0) = v′(L) = 0

then

‖u− uh‖0,2 ≤ C supv∈Sv 6=0

(u− uh |−(pv′)′ + qv)0

|v|2,2


where we recall that S = v /v ∈ H2(0, L), v(0) = v′(L) = 0.Noting that (u− uh |−(pv′)′ + qv)0 = a(u − uh, v) and that a(u − uh, vh) =

0, ∀vh ∈ Vh we have:

‖u− uh‖0,2 ≤ C supv∈Sv 6=0

a(u− uh, v − vh)

|v|2,2

,∀vh ∈ Vh,

or yet

‖u− uh‖0,2 ≤ CM supv∈Sv 6=0

‖u− uh‖1,2 ‖v − vh‖1,2

|v|2,2

,∀vh ∈ Vh.

Taking vh = Ih(v) and recalling (3.12) we obtain

‖u− uh‖0,2 ≤ ‖u− uh‖1,2

CM

h

π

[1 +

h2

π2

] 12

(3.14)

Summarizing all the above estimates we have the following theorems:

Theorem 3.5 Assume that the solution of (P2) belongs to H2(0, L). Then we

have

‖u− uh‖1,2 ≤√

M

µ

h

π

[1 +

h2

π2

] 12

|u|2,2 .

Proof. This is an immediate consequence of the estimates (3.11) and (3.12). q.e.d.

¥

Theorem 3.6 Assume that the solution of (P2) belongs to H2(0, L) for any right

hand side in L2(0, L). Then if f ∈ L2(0, L) the solutions u of (P1) and uh of

(3.10) satisfy

‖u− uh‖0,2 ≤ CM

√M

µ

h2

π2

(1 +

h2

π2

)|u|2,2

where C is the constant of inequality (3.13).

Proof. This is a direct consequence of Theorem 3.5, and of the estimate (3.14).

q.e.d. ¥Let us now consider the case where the right hand side of (P2) is given by an

arbitrary bounded linear functional F over V , or even that p is just a bounded


function, i.e. 0 < α ≤ p(x) ≤ A, ∀x ∈ [0, L]. Under such conditions as we know,

it is not possible to assert that u ∈ H2(0, L). However we still can prove that

uh → u in the norm ‖·‖1,2 as h goes to zero, according to

Theorem 3.7 The solution uh of (3.10) converges to the solution u of (P1) as h

goes to zero, under Assumptions 1 and 2.

Proof. We know that S is dense in V . Hence ∀ε > 0 there exists uε ∈ S such

that ‖u− uε‖1,2 < ε/2.

Now we may use the interpolate Ih(uε) ∈ Vh of uε, for which we have:

‖uε − Ih(uε)‖1,2 ≤ Ch |uε|2,2 , C =(1 + L2/π2)

12

π.

Hence for the given ε we may choose h small enough to have

‖uε − Ih(uε)‖1,2 <ε

2.

Therefore infv∈Vh

‖u− v‖1,2 ≤ ‖u− Ih(uε)‖1,2 < ε, provided h is small enough, for

any ε > 0, and the result follows from bound (3.11). q.e.d. ¥

Theorem 3.8 Assume that |p′| is bounded. Then there exists a constant C0 in-

dependent of h such that the solution of (P2) and (3.10) satisfy

‖u− uh‖0,2 ≤ C0h |u|1,2 . (3.15)

Proof. First we note that since |p′| is bounded the solution of (P1) belongs

H2(0, L) for every right hand side in L2(0, L). Hence we may apply Nitsche’s

trick and (3.14) holds.

On the other hand we have

‖u− uh‖21,2 ≤

M

µ‖u− Πh(u)‖2

1,2 ≤M

µ

[‖u− Πh(u)‖2

1,2 + ‖Πh(u)‖21,2

].

Thus from the Pythagorean Theorem we obtain

‖u− uh‖21,2 ≤

M

µ‖u‖2

1,2 ≤M

µC21

|u|21,2 , (3.16)


where C1 is the constant specified in Proposition 2.1. Combining (3.14) and (3.16)

we establish (3.15) with

C0 =CM

πC1

√M

µ

(1 +

L2

π2

).

q.e.d. ¥

Remark 3.9 Of course if the right hand side of (P1) belongs to L2(0, L), under

the assumptions of Theorem 3.8 estimate (3.15) is suboptimal, since in this case

we may rather apply Theorem 3.6. ¥

3.4 More General Cases and Higher Order Methods

We conclude this chapter by considering other types of finite element methods

applied not only to model problem (P1), but also to some important variants of

this equation, such as:

For 0 < α ≤ p ≤ A and 0 ≤ q ≤ B:

−(pu′)′ + qu = f

u(0) = u(L) = 0(The Dirichlet problem) (3.17)

For 0 < α ≤ p ≤ A and 0 ≤ β ≤ q ≤ B:

−(pu′)′ + qu = f

(pu′)(0) = (pu′)(L) = 0(The Neumann problem) (3.18)

and

−(pu′)′ + qu = f

γu(0)− (pu′)(0) = 0

γu(L) + (pu′)(L) = 0

γ > 0

(The Fourier problem) (3.19)

3.4 More General Cases and Higher Order Methods 53

The reader may easily establish the equivalence between these three problems

under the assumption f ∈ L2(0, L), respectively with

Find u ∈ H10(0, L) such that∫ L

0

pu′v′dx +

∫ L

0

quvdx =

∫ L

0

fvdx

where H10(0, L) = v /v ∈ H1(0, L), v(0) = v(L) = 0

(3.20)


0

pu′v′dx +

∫ L

0

quvdx =

∫ L

0

fvdx, ∀v ∈ H1(0, L)(3.21)


0

pu′v′dx +

∫ L

0

quvdx + γ [u(0)v(0) + u(L)v(L)] =

∫ L

0

fvdx, ∀v ∈ H1(0, L)

(3.22)

He or she may also demonstrate that all those problems have a unique solution

under the respective assumptions on p, q and γ.

Whatever the case one may approximate u by means of the Ritz method, and in

particular by finite element methods such as the already presented piecewise linear

method. Actually in this particular case Vh is defined in the following manner:

Vh = span ϕiN−1i=1 for (3.20)

Vh = span ϕiNi=0 for (3.21) and (3.22).

Furthermore there is no essential difficulty to establish that like in the case of

(P2) the approximation uh ∈ Vh of u determined by this finite element method

satisfies, in all the three cases

‖u− uh‖1,2 ≤ C1h |u|2,2 ,

and if |p′| is bounded

‖u− uh‖0,2 ≤ C0h2 |u|2,2

where C1 and C0 are constants independent of u and h.


Let us now consider higher order piecewise polynomial approximations of second

order linear differential equations of the type −(pu′)′ + qu = f in (0, L), with

different boundary conditions leading to well-posed variational problems such as

those given in (P2),(3.20),(3.21) and (3.22).

Denoting by V the corresponding test function space, we define its approxima-

tion space Vh as follows.

For a given integer k > 1:

Vh =v

/v ∈ V, v|Ki

∈ Pk, ∀i = 1, 2, . . . , N

(3.23)

Notice that in all the cases being considered v ∈ Vh requires that v ∈ C0[0, L],

and the fact that v satisfies the boundary conditions imposed to any function in

V . For instance in approximating (3.20) we must have v(0) = v(L) = 0,∀v ∈ Vh,

while v does not need to satisfy any boundary condition in the case of (3.21) and

(3.22).

If we take for instance k = 2, a convenient basis for Vh in the latter case is

ϕi2Ni=0, where:

ϕ2j has the following form for 0 ≤ j ≤ N − 1

whith

and


ϕ2j−1 has the following form for 1 ≤ j ≤ N :

The reader may draw similar pictures for “natural” bases of Vh, constructed

upon similar principles for some values of k greater than or equal to 3.

Once we have defined the basis of Vh we may compute uh ∈ Vh, namely, the

corresponding Ritz approximation of u, thereby obtaining a system of linear equa-

tions A~c = ~f in all similar to the one generated with piecewise linear elements.

For instance, for k = 2, in the case of problem (3.20), A is a (2N − 1)× (2N − 1)

matrix whose coefficients aij are given by aij =

∫ L

0

(pϕ′jϕ

′i + qϕjϕi

)dx and ~f is

the vector of R2N−1 whose i − th component is

∫ L

0

fϕidx, ~c being the vector of

coefficients in the expansion of uh with respect to the basis ϕi2N−1i=1 .

Finally the approximation errors ‖u− uh‖1,2 and ‖u− uh‖0,2 may be estimated

by means of analyses entirely analogous to those carried out in detail for piecewise

linear elements applied to (P2). The following theorems given without proofs

summarize the main results:

Theorem 3.10 Assume that the solution of the problems (P2),(3.20) (3.21),(3.22)

belongs to u ∈ Hk+1(0, L) where ∀k ∈ N,Hk(0, L) =v

/v, v′, . . . , v(k) ∈ L2(0, L)

.


Then if the problem under consideration is approximated by

Find uh ∈ Vh such that

a(uh, v) =

∫ L

0

fvdx, ∀v ∈ Vh

(3.24)

where Vh is given by (3.23), a is the bilinear form

∫ L

0

pu′v′dx +

∫ L

0

quvdx for

problems (P2),(3.20) and (3.21) or the same expression modified by the addition

of γ[u(0)v(0) + u(L)v(L)] for problem (3.22), then there exists a constant Ck

independent of u and h such that

‖u− uh‖1,2 ≤ Ckhk |u|k+1,2

where |u|k+1,2 =∥∥u(k+1)

∥∥0,2

.

Theorem 3.11 If in addition to the assumption of Theorem 3.10 we further as-

sume that p is such that the solution to any problem among (P2),(3.20) (3.21),(3.22)

belongs to H2(0, L) whenever f ∈ L2(0, L), then there exists a constant C ′k inde-

pendent of u and h such that ‖u− uh‖0,2 ≤ C ′kh

k+1 |u|k+1,2.

Remark 3.12 By an analysis completely analogous to the one carried and for

problem (P2) one establishes that the only assumption on p necessary for Theorem

3.11 to hold is the boundedness of |p′|. ¥

To conclude this Chapter we add that both theorems 3.10 and 3.11 extend

with minor modifications to the approximation of non homogeneous problems by

finite element methods constructed upon piecewise polynomials in the manner

described in this section and in the previous one, such as:

−(pu′)′ + qu = f in (0, L)

u(0) = a1, u(L) = a2

(Non homogeneous Dirichlet problem)

−(pu′)′ + qu = f in (0, L)

(pu′)(0) = b1, (pu′)(L) = b2

(Non homogeneous Neumann problem)


−(pu′)′ + qu = f in (0, L)

γu(0)− (pu′)(0) = c1,

γu(L)− (pu′)(0) = c2

(Non homogeneous Fourier problem)

−(pu′)′ + qu = f in (0, L)

u(0) = a1, (pu′)(L) = b2

(Non homogeneous mixed problem)

among several other possible combinations of the above boundary conditions.

The way to deal with different types of non homogeneous boundary conditions

in a variational context should be exploited by means of exercises proposed as a

complement to these notes.


4Extension to Multi-dimensional Problems

To conclude these lectures we present some main aspects of the application of

the finite element method to solve boundary value partial differential equations

defined in a bounded domain Ω ∈ Rn, with n ≥ 2. We take as a model two second

order linear equations, which we set in variational form before discretizing them

by means of the Ritz method based on continuous functions, whose restriction to

each element of a suitable partition of Ω are constructed upon polynomial bases.

This approach gives rise to the so-called Lagrange finite element method, whose

one-dimensional version was introduced in Chapter 3.

4.1 Sobolev Spaces Hm(Ω)

Let Ω be a bounded open set of Rn, that is, ∃C > 0 such that ∀~x = (x1, x2, . . . ,

xn) ∈ Ω, |~x| ≤ C, where |~x| =[

n∑i=1

x2i

].

We consider the spaceR(Ω) of functions of ~x ∈ Ω, which are Riemann integrable

in Ω. Let R(Ω) be the space of equivalent classes of functions which only differ

60 4. Extension to Multi-dimensional Problems

from eachother in a subset of Ω having a measure equal to zero, that is, a nullset

of Ω(1).

Recalling that every function of R(Ω) is bounded and continuous in Ω, except

eventually at points forming a null set, we may define the following inner product

for R(Ω):

([u]| [v]) =

∫

Ω

u(~x)v(~x)dΩ,

where [v] denotes the equivalence class of v, with corresponding norm ‖[v]‖0,2 =

([v]| [v])120 .

From now on we drop the brackets for simplicity, by (abusively) identifying the

class [v] ∈ R(Ω) with v itself (in so doing we identity R(Ω) with R(Ω)). Now we

define the space L2(Ω) as the completion of R(Ω) with respect to the ‖·‖0,2 norm.

By definition L2(Ω) is then a complete space for this norm, more precisely, the

space of function whose squares are Lebesgue-integrable functions in Ω (cf. [10],

Lebesgue Integration and Measure, ’73).

We further introduce the following notation for the partial derivatives of a

function v defined in Ω: let α = (α1, α2, . . . , αn) be an integer multiindex, i.e.,

αi ∈ N, αi ≥ 0,∀i ∈ 1, 2, . . . , n, and |α| =n∑

i=1

αi. We denote by ∂αv the partial

derivative2 of v of order |α| given by:

∂αv =∂|α|v

∂xα11 ∂xα2

2 . . . ∂xαnn

By definition we set ∂αv = v for α = (0, 0, . . . , 0).

Then for m ∈ N we define the Sobolev space Hm(Ω) for m ∈ N, by:

Hm(Ω) =v

/∂αv ∈ L2(Ω), ∀α such that |α| ≤ m

.

1More precisely a nullset of Ω is a subset of Ω that does not contain any open ball of Ω, that is, a set of the

form B(~x, ε) = ~y /|~x− ~y| < ε, for certain ε ∈ R, ε > 0, and ~x ∈ Ω (which incidentally is not a nullset).2Rigorously we should give a precise sense to such partial derivatives, since we do not necessarily mean usual

ones defined at every point of Ω, that is, strong partial derivatives. Actually when dealing with Sobolev spaces

we have to work with the so-called weak derivatives, i.e., derivatives in the sense of distributions. However it

turns out that the latter coincide with the former almost everywhere for functions in Hm(Ω) (cf. [1], Sobolev

Spaces,’75 )

4.1 Sobolev Spaces Hm(Ω) 61

One may easily establish that the mapping (·| ·)m : Hm(Ω)×Hm(Ω) → R given

by

(u| v)m =∑

|α|≤m

∫

Ω

∂αu∂αvdΩ

defines an inner product in Hm(Ω).

Remark 4.1 Unlike the case of R(Ω) condition (I) holds in the strict sense for

Hm(Ω), since ∂αv = 0 a. e. in Ω, ∀α with |α| = 1 implies that v is constant and

this constant must be zero since

∫

Ω

v2dΩ = 0. ¥

We denote the norm associated with (·| ·)m by ‖·‖m,2 and by |·|m,2 the seminorm

given by |·|2m,2 =∑

|α|≤m

∫

Ω

|∂α(·)|2 dΩ.

We also denote the usual Euclidian inner product of two vectors ~x and ~y of Rn

by ~x · ~y. In so doing we have in particular for H1(Ω):

(u| v)1 =

∫

Ω

(uv + ~grad u · ~grad v

)dΩ

‖v‖1,2 =

[∫

Ω

(|v|2 +

∣∣∣ ~grad v∣∣∣2)

dΩ

] 12

and

|v|1,2 =

[∫

Ω

(∣∣∣ ~grad v∣∣∣2)

dΩ

] 12

.

It is possible to prove that Hm(Ω) is a Hilbert space when equipped with inner

product (·| ·)m.

Let now Ω be the closure of Ω and Γ be the boundary of Ω. If Γ is “sufficiently

smooth”3, by extending functions of H1(Ω) to Γ in the natural way, we may define

its closed subspace H10(Ω), namely

H10(Ω) =

v

/v ∈ H1(Ω), v = 0 on Γ

.

3Here this expression rules out any domain situated on both sides of a portion of its boundary such as the

set Ω =~x ∈ R2 /|~x| < 1, x1 6= x2 for x1 ≥ 0

.


Such “natural” extension is actually the so-called trace of the function v ∈H1(Ω) on Γ. It has an obvious definition in the case of bounded continuous func-

tions in Ω. However, unlike functions of H1(0, L), those in H1(Ω) for Ω ⊂ Rn with

n ≥ 2 need to be neither continuous nor bounded. To understand this, one may

take for instance Ω =~x ∈ R2

/|~x| < 12

and v(~x) = |log |~x|| k2 with k ∈ (0, 1): v

does belong to H1(Ω) but is singular at the origin.

Nevertheless spaces of smooth functions such as C1(Ω) are dense in H1(Ω),

and we may then define the extension of a function v ∈ H1(Ω) to Γ as the limit

in an appropriate sense of the restrictions to Γ of functions forming a sequence

contained in such spaces, that converges to v in the norm of H1(Ω).

Similarly to the case of H10(0, L), the following Friedrichs-Poincare inequality

holds for H10(Ω), that is:

∃Cp(Ω) > 0 such that ‖v‖0,2 ≤ Cp(Ω) |v|1,2 ,∀v ∈ H10(Ω) (4.1)

This implies in particular that |·|1,2 is a norm equivalent to the standard norm

‖·‖1,2 over H10(Ω).

As a consequence, H10(Ω) equipped with the inner product

((u| v))1 =

∫

Ω

~grad u · ~grad vdΩ

is a Hilbert space.

4.2 Green’s Formulae

In order to study variational formulations of partial differential equations, the

following result, giving rise to a series of integration formulae known as Green’s

formulae, is fundamental:

Let ~x ∈ Γ and ~ν(~x) = (ν1, ν2, . . . , νn) be the unit vector normal to Γ and

directed outwards Ω..

4.2 Green’s Formulae 63

If Γ is sufficiently smooth we have∫

Ω

∂v

∂xi

dΩ =

∮

Γ

vνidΓ, i = 1, 2, . . . , n (4.2)

where the “trace” of v on Γ is defined in the sense

described in the previous section.

From relation (4.2) we easily derive the following

results (for sufficiently smooth Γ).

♦ If ~v ∈ [H1(Ω)]n, for div ~v =

n∑i=1

∂vi

∂xi

we have:

∫

Ω

div ~vdΩ =

∮

Γ

~v · ~νdΓ. (4.3)

♦ If ~u ∈ [H1(Ω)]n

and q ∈ H1(Ω) we have:

∫

Ω

~u · ~grad qdΩ =

∮

Γ

q~u · ~vdΓ−∫

Ω

qdiv ~udΩ (4.4)

♦ Denoting by∂u

∂νthe normal derivative of u ∈ H2(Ω) on Γ given by ~grad u·~ν,

for v ∈ H1(Ω) the so-called First Green’s formula holds:

∫

Ω

~grad u · ~grad vdΩ =

∮

Γ

∂u

∂νvdΓ−

∫

Ω

∆uvdΩ (4.5)

where ∆u is the laplacian of u, namely, ∆u =n∑

i=1

∂2u

∂x2i

.

While (4.3) is a trivial consequence of (4.2),(4.4) follows from (4.3) by taking

~v = q~u, which yields div ~v = ~u · ~grad q + qdiv ~u. Finally (4.5) results from (4.4)

by taking ~u = ~grad u, q = v, and noting that div ~grad u = ∆u.

Remark 4.2 The fact that all the above integrals make sense under the assump-

tion that the different functions and fields involved belong to the given spaces,

follows from the Schwartz inequality. Indeed it implies that the product of two


square integrable functions in a bounded domain is integrable in this domain. Ob-

serve also the further assumption that all the functions involved in integrals on Γ

are square integrable (on Γ), which implies that all the corresponding integrands

are integrable on Γ, by the same reason. In particular the trace of a function of

H1(Ω) is square integrable on Γ, which implies that the trace of∂v

∂xi

on Γ is also

square integrable, provided v ∈ H2(Ω), since ∀i, ∂v

∂xi

∈ H1(Ω) in this case. For

more details we refer to [1], Sobolev Spaces,’75. ¥

4.3 Model Variational Problems in N-dimensional Spaces

Let us now consider a model, two variational problems corresponding to linear

elliptic partial differential equations in an n-dimensional bounded domain Ω, for

n = 2 or n = 3, namely:

1. The homogeneous Dirichlet problem for the −∆ operator:

Here V = H10(Ω) and f ∈ L2(Ω). We wish to solve (3.1) with

a(u, v) =

∫

Ω

~grad u · ~grad vdΩ

F (v) =

∫

Ω

fvdΩ.

In this case Theorem 3.1 of the previous Chapter directly applies, since

a(·, ·) is nothing but the inner product ((·| ·))1 for which H10(Ω) is a Hilbert

space. Moreover

|F (v)| ≤ ‖f‖0,2 ‖v‖0,2 ≤ CF |v|1,2 ,∀v ∈ H10(Ω) with CF = Cp(Ω) ‖f‖0,2 .

Let us now show that in this case problem (3.1) is equivalent to the partial

differential equation −∆u = f a.e. in Ω

u = 0 a.e. on Γ(4.6)

4.3 Model Variational Problems in N-dimensional Spaces 65

Although this is by no means essential, we will do it by assuming that the

solution u to the variational problem belongs to H2(Ω), which is always true

if Ω is convex.

In this case we may apply (4.5), thereby obtaining:

a(u, v) =

∮

Γ

∂u

∂νvdΓ−

∫

Ω

∆uvdΩ = F (v) =

∫

Ω

fvdΩ, ∀v ∈ H10(Ω).

Since the integral on Γ vanishes, we have:

∫

Ω

(−∆u− f)vdΩ = 0, ∀v ∈ H10(Ω).

Here we may use the following density result:

H10(Ω) is dense in L2(Ω) equipped with the norm ‖·‖0,2 .

Although this assertion would not be surprising if H10(Ω) were replaced with

H1(Ω), this density result can be established by means of arguments much

similar to those employed to prove that S = V in Chapter 2.

Now since ∆u + f ∈ L2(Ω),∀v ∈ L2(Ω) we may take vn ⊂ H10(Ω) such

that ‖vn − v‖0,2 → 0. This yields (∀v ∈ L2(Ω))

∫

Ω

(∆u + f)vdΩ =

∫

Ω

(∆u + f)(v − vn)dΩ ≤ ‖∆u + f‖0,2 ‖v − vn‖0,2

and hence ∫

Ω

(∆u + f)vdΩ = 0,∀v ∈ L2(Ω),

which implies that ∆u + f = 0 a.e. in Ω.

Conversely, assuming that the solution of (4.6) belongs to H2(Ω), we may

rebuild the original variational problem the other way around, by multi-

plying both sides of the differential equation with an arbitrary v ∈ H10(Ω),

integrating the result over Ω, and then applying first Green’s formula. In so

doing the former is seen to be equivalent to the latter, at least as long as

we can assert that u ∈ H2(Ω).


2. The homogeneous Neumann problem for the operator −∆ + I (I being the

identity operator):

Let V = H1(Ω), f ∈ L2(Ω) and (3.1) be such that

a(u, v) =

∫

Ω

[~grad u · ~grad v + uv

]dΩ

F (v) =

∫

Ω

fvdΩ.

Here again Theorem 3.1 of Chapter 3 applies directly, as a(·, ·) is nothing but

the standard inner product of H1(Ω) and F (v) ≤ CF ‖v‖1,2 ,∀v ∈ H1(Ω),

with CF = ‖f‖0,2.

In the same manner as for the Dirichlet problem, we may show that, pro-

vided the solution to the so-defined variational problem belongs to H2(Ω)

- which is true here again at least if Ω is convex - then u is the solution of

the partial differential equation

−∆u + u = f a.e. in Ω∂u

∂ν= 0 a.e. on Γ

(4.7)

The reader may verify that, conversely, any solution of (4.7) belonging to

H2(Ω) is a solution of the variational problem

a(u, v) ≡ (u| v)1 = (f | v)0 ≡ F (v),∀v ∈ H1(Ω).

Remark 4.3 The equivalence of both equations (4.6) and (4.7) with the corre-

sponding problem (3.1) holds under assumption much weaker than u ∈ H2(Ω).

However in order to establish it for much less smooth boundaries Γ, than for con-

vex domains, one must use some regularity results for elliptic partial differential

equations that lie beyond the scope of these notes. ¥

As a final point addressed in this section we recall that, according to Theorem

3.2 of the previous Chapter, the variational problem of the two examples we just

considered, are respectively equivalent to the following minimization problems:

4.4 Lagrange Finite Element Solution Methods 67

1. Find u ∈ H10(Ω) such that ∀v ∈ H1

0(Ω)

1

2

∫

Ω

∣∣∣ ~grad u∣∣∣2

dΩ−∫

Ω

fudΩ ≤ 1

2

∫

Ω

∣∣∣ ~grad v∣∣∣2

dΩ−∫

Ω

fvdΩ;

2. Find u ∈ H1(Ω) such that ∀v ∈ H1(Ω)

1

2

[∫

Ω

∣∣∣ ~grad u∣∣∣2

+ u2

]dΩ−

∫

Ω

fudΩ ≤ 1

2

∫

Ω

[∣∣∣ ~grad v∣∣∣2

+ v2

]dΩ−

∫

Ω

fvdΩ.

4.4 Lagrange Finite Element Solution Methods

The Ritz method is again a suitable numerical approach to solve the model

problems considered in the previous sections, as it is for a very large spectrum of

similar boundary value problems for partial differential equations.

Recalling the main principles of this technique described in the previous Chap-

ter, we go directly into the description of the particular finite dimensional spaces

VN with corresponding bases, associated with the most classical and popular so-

lution method os this class: the Lagrange finite element method.

The Ritz method corresponds to the choice of particular spaces VN , namely,

N−dimensional spaces Vh consisting of continuous functions whose restriction

to every element of a partition of Ω into subset of a given shape fulfilling some

compatibility conditions at the interfaces, are functions of a given type as well.

Vh should possibly be a subspace of the space of test functions V , i.e., H10(Ω) or

H1(Ω). Since in the former case it is not so easy to construct Vh ⊂ V in general, in

order to simplify the presentation we shall assume henceforth that Ω is a polygon

for n = 2 and a polyhedron for n = 3.

In this case we consider only the following types of partitions:

• Th consists of closed sets T which are triangles if n = 2 and tetrahedrons if

n = 3;

• The intersection of two elements of Th is either empty, a common vertex, a

common edge, or a common face (for n = 3 only);


•⋃

T∈Th

= Ω.

The above conditions imply that ∀T ∈ Th, T ∩ Γ is either the empty set or a

subset of the set of vertices, edges or faces (for n = 3 only) of T .

In so doing we define for the Neumann problem:

Vh =v

/v|T ∈ Pk,∀T ∈ Th, v ∈ C0(Ω)

and for the Dirichlet problem we take the intersection of the above space with

H10(Ω), whereby Pk denotes the space of polynomials in n variables of degree less

than or equal to k.

In order to illustrate how one may compute with such spaces, we consider the

case K = 1: letting SiKi=1 be the set of vertices of the elements of Th numbered

in such a way that the last L vertices are those belonging to Γ a natural choice

for the basis B of Vh, according to the case, is:

√For the Dirichlet problem: B = ϕiN

i=1 with N = K − L.

√For the Neumann problem: B = ϕiN

i=1 with N = K,

Where ϕi(Sj) = δij for 1 ≤ i, j ≤ N .

Once we have defined an appropriate basis B = ϕiNi=1 for Vh, in the general

case, with N = dimVh, we are led to a system of equations for the components

c1, c2, , cN of uh ∈ Vh with respect to B, namely, the solution of:

Find uh ∈ Vh such that

a(uh, v) = F (v),∀v ∈ Vh

(4.8)

This system equivalent to (4.8) is again A~c = ~f , where aij = a(ϕj, ϕi), ~c =

(c1, c2, . . . , cN) and fi = F (ϕi), 1 ≤ i, j ≤ N .

Now according to Theorem 3.1 of Chapter 3, the following error bound applies

to both model problems:

∃C > 0 such that ‖u− uh‖1,2 ≤ C infv∈Vh

‖u− v‖1,2


where C = 1 in the case of the Neumann problem, and C =[1 + C2

p(Ω)]− 1

2 in the

case of the Dirichlet problem.

Whenever u is continuous and bounded , one may conveniently bound above the

infimum term by ‖u− Ih(u)‖1,2, where Ih(u) is the function of Vh that interpolates

u at the nodes of the partition Th associated with the basis functions of the

Lagrange finite element method (for instance in the case of piecewise linear finite

elements, these are the vertices of the triangles or of the tetrahedrons of Th).

This choice is not always possible for n ≥ 2, since a function of H1(Ω) is not

necessarily bounded or continuous, and hence this interpolate might simply not

exist. However here we may apply the Sobolev Embedding Theorem (cf. [1], Sobolev

Spaces, ’75 ) which states that whenever u ∈ H2(Ω), then u ∈ C0(Ω) ∩ B(Ω) for

n = 2 and n = 3, where B(Ω) is the space of bounded functions in Ω. Actually

since we are only considering polygonal or polyhedral domains we always have

u ∈ H2(Ω) at least when Ω is convex, as previously stated.

Anyway, provided u ∈ H2(Ω) we have

‖u− uh‖1,2 ≤ C ‖u− Ih(u)‖1,2 .

Let now hT = max~x,~y∈T

|~x− ~y|, and h = maxT∈T

hT . We also define ρT to be the

radius of the circle or sphere inscribed in T ∈ Th (that is ρT = maxB(~x,ρ)⊂T

ρ). We

shall make the following assumption on an infinite family of partitionThh that

we supposedly employ to solve a corresponding sequence of problems (4.8) with

h ≤ h0,∀Th.

∃ a constant c > 0 such thatρT

hT

≥ c, ∀T ∈ Th and ∀h. (4.9)

Such family of partitions is called quasiuniform.

Then we can prove the following error estimate for both model problems con-

sidered in this Chapter: If u ∈ Hk+1(Ω) there exists a constant CM independent

of u and h such that:

‖u− uh‖1,2 ≤ CMhk |u|k+1,2 . (4.10)


Estimate implies in particular that if the quasiuniform family of partitions

Thh is such that h → 0, then uh converges to u in H1(Ω) as h goes to zero.

Moreover (4.10) also implies that uh is the exact solution u,∀h, whenever u is a

polynomial of degree less than or equal to k in Ω.

At this point we illustrate the importance of the assumption that Thh be a

quasiuniform through the above example given by [9] ’73:

Take n = 2 and let the triangle T with vertices (0, h2), (0,−h

2) and (λ, 0) be

an element of Th. Letting λ = h2 we have hT = h and (4.9) does not hold, for:

ρT = 14

(√1 + 4h2 − 1

)= h2/2 +O(h3).

Now consider the function v(x1, x2) = x22 and piecewise linear elements. The re-

striction of Ih(v) to T is the linear function IT v such that IT v(0, h2) = IT v(0,−h

2) =

h2/4 and IT v(λ, 0) = 0.

We have∂(IT v)

∂x1

= −1/4 and∂(IT v)

∂x2

= 0, whereas∂v

∂x1

= 0 and∂v

∂x1

= 2x2.

This yields |v − IT v|1,2,T

def=

[∫

T

∣∣∣ ~grad (v − IT v)∣∣∣2

dx1dx2

] 12

≥[∫

T

(−1

4)2dx1dx2

] 12

=

√area(T )

4, while |v|2,2,T

def=

∑

|α|=2

∫

T

|∂αv|2 dx1dx2

12

=

∑

|α|=2

∫

T

22dx1dx2

12

≤

8 |v − IT v|1,2,T . This indicates that

‖u− Ih(u)‖1,2 ≥[∑

T∈T|u− IT u|21,2,T

] 12

can be no means be bounded above by a term of the form CMh |u|2,2 with CM

independent of u and h!

To conclude we give an estimate of the error measured in the ‖·‖0,2 norm, in the

case where Ω is convex, so that the solutions of the partial differential equations

(4.6) and (4.7) belong to H2(Ω) for every f ∈ L2(Ω).

Here again we may use Nitsche’s trick and establish that there exists a constant

C ′M independent of u and h such that

‖u− uh‖0,2 ≤ C ′Mhk+1 |u|k+1,2 (4.11)


provided u ∈ Hk+1(Ω).

The proof of error estimate (4.10) requires in general fairly more sophisticated

tools than those employed for one dimensional problems. Nevertheless they are

definitively analogous and qualitatively comparable to those given in Chapter 3.

For further details we refer to several available works on the subject, among which

CIARLET [2] certainly lies among the most celebrated ones.


References

[1] ADAMS, R. A., Sobolev Spaces, Academic Press, New York, 1975.

[2] CIARLET, P. G., The Finite Element Method for Elliptic Problems, North

Holland, Amsterdam, 1976.

[3] COLLATZ, L., Funktionalanalysis und Numerische Mathematik, Springer

Verlag, Berlim, 1964.

[4] COURANT, R. & HILBERT, D., Methods of Mathematical Physics, John

Wiley, N. Y., 1962.

[5] DAUTRAY, R. & LIONS, J. L., Mathematical Analysis and Numerical Meth-

ods for Science and Technology, Springer Verlag, 1993.

[6] LUENBERGER, D. G., Optimization by Vector Space Methods, John Wiley

& Sons Inc., New York, 1969.

[7] MIKHLIN, S. G., Variational Methods in Mathematical Physics, Moscow,

1957.

74 References

[8] SCHWARTZ, L., Theorie des Distributions, Hermann et Cie. editions, Paris,

1950.

[9] STRANG, G. & FIX, G. J., An Analysis of the Finite Element Method,

Prentice Hall, Englewood Cliffs, N. J., 1973.

[10] WEIR,A. J., Lebesgue Integration and Measure, Cambridge University Press,

Cambridge, 1974

[11] Ruas, V., Uma Introducao aos Problemas Variacionais, Editora Guanabara

Dois S. A., Rio de Janeiro, 1979.

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An Introduction to the Mathematical Foundations of the ...€¦ · element method does, it is readily seen that continuous functions may be used for (P2) whereas continuously diﬀerentiable