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-