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is a bilinear form, i.e. it is linear with respect to both arguments u and
v. Problem (12.43) is called the weak formulation of problem (12.1). Since
(12.43) contains only the ¬rst derivative of u it might cover cases in which
a classical solution u ∈ C 2 ([0, 1]) of (12.41) does not exist although the
physical problem is well de¬ned.
If for instance, ± = 1, β = γ = 0, the solution u(x) denotes of the
displacement at point x of an elastic cord having linear density equal to f ,
whose position at rest is u(x) = 0 for all x ∈ [0, 1] and which remains ¬xed
at the endpoints x = 0 and x = 1. Figure 12.2 (right) shows the solution
u(x) corresponding to a function f which is discontinuous (see Figure 12.2,
left). Clearly, u does not exist at the points x = 0.4 and x = 0.6 where f
is discontinuous.
If (12.41) is supplied with non homogeneous boundary conditions, say
u(0) = u0 , u(1) = u1 , we can still obtain a formulation like (12.43) by
proceeding as follows. Let u(x) = xu1 + (1 ’ x)u0 be the straight line that
¯
0 0
interpolates the data at the endpoints, and set u= u(x) ’ u(x). Then u∈ V
¯
546 12. Two-Point Boundary Value Problems

0

’0.005

f(x) ’0.01


u(x)
’0.015

0.6
0.4 1x
0 ’0.02

’0.025

’0.03

’0.035

’1 ’0.04

’0.045

’0.05
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1




FIGURE 12.2. Elastic cord ¬xed at the endpoints and subject to a discontinuous
load f (left). The vertical displacement u is shown on the right

satis¬es the following problem
0 0
¬nd u∈ V : a(u, v) = (f, v) ’ a(¯, v) for all v ∈ V.
u

A similar problem is obtained in the case of Neumann boundary conditions,
say u (0) = u (1) = 0. Proceeding as we did to obtain (12.43), we see
that the solution u of this homogeneous Neumann problem satis¬es the
same problem (12.43) provided the space V is now H1 (0, 1). More general
boundary conditions of mixed type can be considered as well (see Exercise
12).


12.4.2 A Quick Introduction to Distributions
Let X be a Banach space, i.e., a normed and complete vector space. We
say that a functional T : X ’ R is continuous if limx’x0 T (x) = T (x0 )
for all x0 ∈ X and linear if T (x + y) = T (x) + T (y) for any x, y ∈ X and
T (»x) = »T (x) for any x ∈ X and » ∈ R.
Usually, a linear continuous functional is denoted by T, x and the sym-
bol ·, · is called duality. As an example, let X = C 0 ([0, 1]) be endowed
with the maximum norm · ∞ and consider on X the two functionals
de¬ned as
1
T, x = x(0), S, x = x(t) sin(t)dt.
0
It is easy to check that both T and S are linear and continuous functionals
on X. The set of all linear continuous functionals on X identi¬es an abstract
space which is called the dual space of X and is denoted by X .

We then introduce the space C0 (0, 1) (or D(0, 1)) of in¬nitely di¬er-
entiable functions having compact support in [0, 1], i.e., vanishing outside
a bounded open set (a, b) ‚ (0, 1) with 0 < a < b < 1. We say that
vn ∈ D(0, 1) converges to v ∈ D(0, 1) if there exists a closed bounded set
K ‚ (0, 1) such that vn vanishes outside K for each n and for any k ≥ 0
(k)
the derivative vn converges to v (k) uniformly in (0, 1).
12.4 The Galerkin Method 547

The space of linear functionals on D(0, 1) which are continuous with
respect to the convergence introduced above is denoted by D (0, 1) (the
dual space of D(0, 1)) and its elements are called distributions.
We are now in position to introduce the derivative of a distribution. Let
T be a distribution, i.e. an element of D (0, 1). Then, for any k ≥ 0, T (k)
is also a distribution, de¬ned as

∀• ∈ D(0, 1).
T (k) , • = (’1)k T, •(k) , (12.45)

As an example, consider the Heaviside function

x ≥ 0,
1
H(x) =
0 x < 0.

The distributional derivative of H is the Dirac mass δ at the origin, de¬ned
as
v ’ δ(v) = v(0), v ∈ D(R).
From the de¬nition (12.45), it turns out that any distribution is in¬nitely
di¬erentiable; moreover, if T is a di¬erentiable function its distributional
derivative coincides with the usual one.


12.4.3 Formulation and Properties of the Galerkin Method
Unlike the ¬nite di¬erence method which stems directly from the di¬er-
ential (or strong) form (12.41), the Galerkin method is based on the weak
formulation (12.43). If Vh is a ¬nite dimensional vector subspace of V , the
Galerkin method consists of approximating (12.43) by the problem

¬nd uh ∈ Vh : a(uh , vh ) = (f, vh ) ∀vh ∈ Vh . (12.46)

This is a ¬nite dimensional problem. Actually, let {•1 , . . . , •N } denote a
basis of Vh , i.e. a set of N linearly independent functions of Vh . Then we
can write
N
uh (x) = uj •j (x).
j=1


The integer N denotes the dimension of the vector space Vh . Taking vh = •i
in (12.46), it turns out that the Galerkin problem (12.46) is equivalent to
seeking N unknown coe¬cients {u1 , . . . , uN } such that

N
∀i = 1, . . . , N.
uj a(•j , •i ) = (f, •i ) (12.47)
j=1
548 12. Two-Point Boundary Value Problems

We have used the linearity of a(·, ·) with respect to its ¬rst argument, i.e.
N N
a( uj •j , •i ) = uj a(•j , •i ).
j=1 j=1

If we introduce the matrix AG = (aij ), aij = a(•j , •i ) (called the sti¬ness
matrix), the unknown vector u = (u1 , . . . , uN ) and the right-hand side
vector fG = (f1 , . . . , fN ), with fi = (f, •i ), we see that (12.47) is equivalent
to the linear system
AG u = fG . (12.48)
The structure of AG , as well as the degree of accuracy of uh , depends on
the form of the basis functions {•i }, and therefore on the choice of Vh .
We will see two remarkable instances, the ¬nite element method, where
Vh is a space of piecewise polynomials over subintervals of [0, 1] of length not
greater than h which are continuous and vanish at the endpoints x = 0 and
1, and the spectral method in which Vh is a space of algebraic polynomials
still vanishing at the endpoints x = 0, 1.
However, before speci¬cally addressing those cases, we state a couple of
general results that hold for any Galerkin problem (12.46).


12.4.4 Analysis of the Galerkin Method
We endow the space H1 (0, 1) with the following norm
0
±1 1/2
 
|v|H1 (0,1) = |v (x)| dx
2
. (12.49)
 
0

We will address the special case where β = 0 and γ(x) ≥ 0. In the most
general case given by the di¬erential problem (12.41) we shall assume that
the coe¬cients satisfy
1
’ β + γ ≥ 0, ∀x ∈ [0, 1]. (12.50)
2
This ensures that the Galerkin problem (12.46) admits a unique solution
depending continuously on the data. Taking vh = uh in (12.46) we obtain
1 1

±0 |uh |2 1 (0,1) ¤ γuh uh dx = (f, uh ) ¤ f
±uh uh dx + uh L2 (0,1) ,
L2 (0,1)
H
0 0

where we have used the Cauchy-Schwarz inequality (8.29) to set the right-
hand side inequality. Owing to the Poincar´ inequality (12.16) we conclude
e
that
CP
|uh |H1 (0,1) ¤ f L2 (0,1) . (12.51)
±0
12.4 The Galerkin Method 549

Thus, the norm of the Galerkin solution remains bounded (uniformly with
respect to the dimension of the subspace Vh ) provided that f ∈ L2 (0, 1).
Inequality (12.51) therefore represents a stability result for the solution of
the Galerkin problem.
As for convergence, we can prove the following result.

’1 2
Theorem 12.3 Let C = ±0 ( ± + CP γ ∞ ); then, we have



|u ’ uh |H1 (0,1) ¤ C min |u ’ wh |H1 (0,1) . (12.52)
wh ∈Vh

Proof. Subtracting (12.46) from (12.43) (where we use vh ∈ Vh ‚ V ), owing to
the bilinearity of the form a(·, ·) we obtain

a(u ’ uh , vh ) = 0 ∀vh ∈ Vh . (12.53)

Then, setting e(x) = u(x) ’ uh (x), we deduce

±0 |e|2 1 (0,1) ¤ a(e, e) = a(e, u ’ wh ) + a(e, wh ’ uh ) ∀wh ∈ Vh .
H


The last term is null due to (12.53). On the other hand, still by the Cauchy-
Schwarz inequality we obtain
1 1

a(e, u ’ wh ) ±e (u ’ wh ) dx + γe(u ’ wh ) dx
=
0 0
¤± ∞ |e|H1 (0,1) |u ’ wh |H1 (0,1) + γ u ’ wh
e L2 (0,1) .
∞ L2 (0,1)


The desired result (12.52) now follows by using again the Poincar´ inequality for
e
both e L2 (0,1) and u ’ wh L2 (0,1) . 3

The previous results can be obtained under more general hypotheses on
problems (12.43) and (12.46). Precisely, we can assume that V is a Hilbert
space, endowed with norm · V , and that the bilinear form a : V — V ’ R
satis¬es the following properties:

∃±0 > 0 : a(v, v) ≥ ±0 v ∀v ∈ V (coercivity),
2
(12.54)
V



∃M > 0 : |a(u, v)| ¤ M u ∀u, v ∈ V (continuity).
v (12.55)
V V


Moreover, the right hand side (f, v) satis¬es the following inequality

|(f, v)| ¤ K v ∀v ∈ V.
V


Then both problems (12.43) and (12.46) admit unique solutions that satisfy

K K
¤ ¤
u , uh .
V V
±0 ±0
550 12. Two-Point Boundary Value Problems

This is a celebrated result which is known as the Lax-Milgram Lemma (for
its proof see, e.g., [QV94]). Besides, the following error inequality holds
M
u ’ uh ¤ min u ’ wh . (12.56)
V V
±0 wh ∈Vh
The proof of this last result, which is known as C´a™s Lemma, is very similar
e
to that of (12.52) and is left to the reader.

We now wish to notice that, under the assumption (12.54), the matrix
introduced in (12.48) is positive de¬nite. To show this, we must check that
vT Bv ≥ 0 ∀v ∈ RN and that vT Bv = 0 ” v = 0 (see Section 1.12).
Let us associate with a generic vector v = (vi ) of RN the function vh =
N
j=1 vj •j ∈ Vh . Since the form a(·, ·) is bilinear and coercive we get

N N N N
T
v AG v = vi aij vj = vi a(•j , •i )vj
« 
j=1 i=1 j=1 i=1
N N N N
a(vj •j , vi •i ) = a  vi •i 
= vj •j ,
j=1 i=1 j=1 i=1
= a(vh , vh ) ≥ ± ≥ 0.
vh 2
V

Moreover, if vT AG v = 0 then also vh 2
= 0 which implies vh = 0 and
V
thus v = 0.

It is also easy to check that the matrix AG is symmetric i¬ the bilinear
form a(·, ·) is symmetric.
For example, in the case of problem (12.41) with β = γ = 0 the ma-
trix AG is symmetric and positive de¬nite (s.p.d.) while if β and γ are
nonvanishing, AG is positive de¬nite only under the assumption (12.50).
If AG is s.p.d. the numerical solution of the linear system (12.48) can be
e¬ciently carried out using direct methods like the Cholesky factorization
(see Section 3.4.2) as well as iterative methods like the conjugate gradient
method (see Section 4.3.4). This is of particular interest in the solution of
boundary value problems in more than one space dimension (see Section
12.6).


12.4.5 The Finite Element Method
The ¬nite element method (FEM) is a special technique for constructing
a subspace Vh in (12.46) based on the piecewise polynomial interpolation
considered in Section 8.3. With this aim, we introduce a partition Th of
[0,1] into n subintervals Ij = [xj , xj+1 ], n ≥ 2, of width hj = xj+1 ’ xj ,
j = 0, . . . , n ’ 1, with
0 = x0 < x1 < . . . < xn’1 < xn = 1
12.4 The Galerkin Method 551

and let h = max(hj ). Since functions in H1 (0, 1) are continuous it makes
0
Th
sense to consider for k ≥ 1 the family of piecewise polynomials Xh intro-
k

duced in (8.22) (where now [a, b] must be replaced by [0, 1]). Any function
vh ∈ Xh is a continuous piecewise polynomial over [0, 1] and its restriction
k

over each interval Ij ∈ Th is a polynomial of degree ¤ k. In the following
we shall mainly deal with the cases k = 1 and k = 2.
Then, we set
k,0
Vh = Xh = vh ∈ Xh : vh (0) = vh (1) = 0 .
k
(12.57)

The dimension N of the ¬nite element space Vh is equal to nk ’ 1. In the
following the two cases k = 1 and k = 2 will be examined.
To assess the accuracy of the Galerkin FEM we ¬rst notice that, thanks
to C´a™s lemma (12.56), we have
e

min u ’ wh ¤ u ’ Πk u (12.58)
H1 (0,1) H1 (0,1)
h
0 0
wh ∈Vh


where Πk u is the interpolant of the exact solution u ∈ V of (12.43) (see
h
Section 8.3). From inequality (12.58) we conclude that the matter of esti-
mating the Galerkin approximation error u ’ uh H1 (0,1) is turned into the
0
estimate of the interpolation error u ’ Πk u H1 (0,1) . When k = 1, using
h 0
(12.56) and (8.27) we obtain

M
u ’ uh ¤ Ch u
H1 (0,1) H2 (0,1)
0
±0

provided that u ∈ H2 (0, 1). This estimate can be extended to the case k > 1
as stated in the following convergence result (for its proof we refer, e.g., to
[QV94], Theorem 6.2.1).

Property 12.1 Let u ∈ H1 (0, 1) be the exact solution of (12.43) and uh ∈
0
Vh its ¬nite element approximation using continuous piecewise polynomials
of degree k ≥ 1. Assume also that u ∈ Hs (0, 1) for some s ≥ 2. Then the
following error estimate holds
M
u ’ uh ¤ Chl u (12.59)
H1 (0,1) Hl+1 (0,1)
0
±0
where l = min(k, s ’ 1). Under the same assumptions, one can also prove
that

u ’ uh ¤ Chl+1 u Hl+1 (0,1) . (12.60)
L2 (0,1)



The estimate (12.59) shows that the Galerkin method is convergent, i.e. the
approximation error tends to zero as h ’ 0 and the order of convergence is
552 12. Two-Point Boundary Value Problems

k. We also see that there is no convenience in increasing the degree k of the
¬nite element approximation if the solution u is not su¬ciently smooth. In
this respect l is called a regularity threshold. The obvious alternative to gain
accuracy in such a case is to reduce the stepzise h. Spectral methods, which
will be considered in Section 12.4.7, instead pursue the opposite strategy
(i.e. increasing the degree k) and are thus ideally suited to approximating
problems with highly smooth solutions.
An interesting situation is that where the exact solution u has the min-

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