Weak Formulation

7/31/2019 Weak Formulation

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Chapter 4

Weak formulation of elliptic problems

Alan Turing is reported as saying that PDEs are made by God, theboundary conditions by the Devil ! The situation has changed, Devilhas changed places...We can say that the main challenges are in the

interfaces, with Devil not far away from them. . . Jacques-Louis Lions (1928-2001)

In this chapter, we consider linear elliptic problems that are commonly found in mechanical andphysical partial differential models. The aim is to introduce the notion of weak formulation that giveaccess to existence and uniqueness results for the solutions and that is well suited for the numericalapproximation of such problems.

Contents

4.1 Dirichlet problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

4.2 Neumann problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

4.3 Abstract variational problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

4.4 Variational approximation of elliptic problems . . . . . . . . . . . . . . . . . . 66

4.5 Application to boundary-value problems in R . . . . . . . . . . . . . . . . . . 71

4.6 Maximum principles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74

4.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75

4.1 Dirichlet problem

Let consider a bounded domain Rd with a piecewise C1 continuous boundary and let denote by the unit outer normal vector to . We first recall a Greens identity.

Lemma 4.1 For every function u H2() and every function v H1(), we have the Greens identity:

(u)v dx =d

i=1

xiuxiv dx

u

v d . (4.1)

53


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54 Chapter 4. Weak formulation of elliptic problems

We consider the following model problem:Given a real-valued function f L2(), find a function u defined in solving

u = f in (4.2)

u = 0 on . (4.3)

Suppose the solution u is sufficiently smooth, for example u H2(). Then, we can multiply each sideof the equation (4.2) by a test function v H10 () and integrate on to obtain:

(u)v dx =

f v d x .

Thanks to Greens identity (4.1) and by noticing that v| = 0, we obtain:

v H10 () ,

di=1

xiu xiv dx =

f v d x . (4.4)

The specific boundary conditions and the Theorem 2.21 yield to conclude that the solution u of theproblem (4.4) is such that:

u H10 ()

and the initial problem is now replaced by the following problem:Given a function f L2(), find a function u H10 () such that identity (4.4) holds.

This is the weak formulation of the Dirichlet problem (4.2).We have seen that any solution u of the Dirichlet problem (4.2), sufficiently smooth, is solution of

the problem (4.4). Conversely, a solution u H10 () is a solution of the problem (4.4) if and only if:

D() ,d

i=1 xiuxi dx = f dx ,

since H10 () is the closure ofD() in H1() (cf. Definition 2.14). Consequently, ifu satisfies the equation

(4.4), then the initial equation u = f is satisfied on in the distributional sense and since f L2(),this equation is verified in L2(), thus almost everywhere in . Finally, u H10 () leads to satisfy theboundary condition according to the trace theorem on the boundary .

4.2 Neumann problem

Let consider an open bounded domain Rd with a piecewise C1 continuous boundary . We nowfocus on the following problem:

Given a function f L

2

(

), find a function u defined on

solving

u + u = f in (4.5)

u

= 0 on . (4.6)

Suppose the solution u H2(). Then, following the same procedure as for the Dirichlet problem, weobtain:

v H1() ,d

i=1

xiu xiv dx +

uv dx =

f v d x . (4.7)


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4.3. Abstract variational problems 55

The original boundary-value problem is then replaced by the following:Given a function f L2(), find a solution u H1() verifying equation (4.7).

Any sufficiently smooth solution u of the Neumann problem is solving the previous problem. Conversely,if u H1() solves the problem (4.7), we have

D() ,

di=1

xiuxi dx +

u dx =

f dx ,

and since u and f are L2() functions, the initial equation u + u = f is satisfied in the distributionalsense, in L2(). When u H2(), we can show, using the Greens formula, that equation 4.7 is equivalentto the initial equation and that:

v H1() ,

u

v d = 0 ,

and, since the trace space on is dense in L2(), we have:

w L2() ,

u

w d = 0 ,

which leads to the Neumann boundary condition in the L2() sense. Any smooth solution of equation(4.7) is thus solving the initial Neumann problem.

4.3 Abstract variational problems

The previous examples fit in a general abstract framework that is well suited for the resolution of a largeclass of boundary value elliptic problems. To this end, we consider:

(i) a Hilbert space V on R, endowed with a norm ,

(ii) a bilinear form a defined and continuous on V V, i.e. such that there exists a constant M > 0

such that:u V , v V , a(u, v) Mu v ,

(iii) a linear form l continuous on V, i.e. an element l of the topological dual space V = L(V, R) of Vendowed with the dual norm

lV = supvVv=0

l(v)

v.

Then, we consider the general variational problem:

(P)Given l V , find u V solving the problem

v V , a(u, v) = l(v)(4.8)

As such, this problem is certainly not well-posed. We will give a sufficient condition on the bilinear forma to have a unique solution. We first introduce a definition.

Definition 4.1 The problem (P) is said to be well-posed if there is a unique solution to this problemand if the following linear stability property is satisfied:

C > 0 , l V , u ClV .

The objective is now to determine if and when a variational problem is well-posed. Two fundamentalresults will help us to solve this question.


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4.3.1 The Lax-Milgram theorem

Definition 4.2 The bilinear form a(, ) is said to be V-elliptic if there exists a constant > 0 such that

v V , a(v, v) v2 .

Theorem 4.1 (Lax-Milgram lemma) Let V be a Hilbert space, a L(V V,R) be a bilinear form,and l V be a linear form continuous on V. Then, under the condition that:

(c0) the bilinear form a is continuous and V-elliptic with constant,

the problem P is well-posed (i.e. has a unique solution) and we have the estimate:

l V , uV 1

lV ,

(i.e. the mapping u l is an isomorphism of V on V).

Proof. Since the linear form l is continuous on V, the Riesz representation theorem asserts the existence of aunique element wl V such that

v V , l(v) = wl, v .

This relation defines a linear one-to-one isometric mapping l wl from V on V. Similarly, u being fixed in V,the linear form v a(u, v) is continuous on V. It is thus an isomorphism. Invoking the representation theoremagain, there exists a unique element w V such that:

v V , a(u, v) = w, v ,

and we pose w = Au. We claim that A : V V is a bounded continuous linear operator. It follows that theproblem (P) is equivalent to finding u V solving

Au = wl .

The existence and uniqueness are obtained by establishing that the linear operator A is a bijective mapping fromV to V. A being continuous, the inverse A1 is continuous from V to V.

(i) A is an injection, i.e. Ker(A) = {0}. Actually, since a is V-elliptic, we have

v V , v2 a(v, v) = Av, v Av v ,

and thus,v V , Av v .

This shows that A is an injection, since:

v Ker(A) Av = 0 Av, v = 0

thus v = 0 and, consequently, Ker(A) = {0}.

(ii) A is a surjection, i.e. ImA = R(A) = V. We need to show that R(A) is closed and dense in V.Let consider w R(A) (the closure of R(A) in V) and let (Avm)mN be a sequence of R(A) that convergesto w in V. We have:

Avm Avn vm vnsuch that (vm) is a Cauchy sequence in the Hilbert space V: it converges to an element v V and (Avm)converges to Av, since A is continuous. Hence, we have:

w = Av R(A) ,

and we can conclude that R(A) is closed in V.

Now, let consider v0 (R(A)), we have

v02 a(v0, v0) = Av0, v0 = 0 ,

anf thus v0 = 0. This shows that (R(A)) is reduced to {0} and thus A is a surjection.


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The linear operator A is a one-to-one mapping from V to V. From the V-ellipticity property, we deduce

v V , A1v 1

v ,

and hence that A1 is continuous. By noticing that:

wl = supvVv=0

wl,v

v= sup

vVv=0

l(v)

v= l ,

we obtain that the solution u to the problem (P) is such that:

u = A1wl 1

wl =

1

lV ,

and the result follows.

When the bilinear form a(, ) is symmetric and positive, the problem (P) can be interpreted as aminimization problem.

Theorem 4.2 Consider a Hilbert space V, a L(V V,R) and l V. Suppose the bilinear form a(, )is symmetric and positive:

u, v V , a(u, v) = a(v, u) and u V , a(u, u) 0 .

Then, u is solution of the problem (P) if and only if u minimizes on V the functional

v V , J(v) =1

2a(v, v) l(v) ,

i.e. J(u) = infvV

J(v).

Proof. Suppose u is solution of the problem (P). A direct calculation gives:

v V , J(v) = J(u) + a(u, v u) l(v u) +1

2a(v u, v u)

= J(u) +1

2a(v u, v u)

J(u) ,

which attests that u minimizes the function J on V. Conversely, let consider v V, ]0, 1] and we posew = u + v. A direct calculation gives:

J(w) J(u) = (a(u, v) l(v)) +2

2a(v, v) 0 .

Dividing by and letting 0 yields

v V , a(u, v) l(v) .

Substituting v by v yields to the equality and thus to conclude that u is solution of the problem (P).

Remark 4.1 (i) When the bilinear form a(, ) is symmetric, the problem (P) corresponds to the min-imization of a quadratic functional on a Hilbert space V, which is the abstract formulation ofnumerous problems in calculus of variations. This explains why(P) is called a variational problem.


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(ii) When the bilinear form a(, ) is symmetric and V-elliptic, the Lax-Milgram theorem indicates thatthe optimization problem infvV J(v) has a unique solution. The V-ellipticity of a can be seen as aproperty of strong convexity of the functional J.

(iii) In many applications, the functional J corresponds to an energy (cf. the deformation of an elastic

membrane).

Example 4.1 We consider the Dirichlet problem (4.2). Its variational (weak) formulation is that ofproblem (P), if we pose

V = H10 () endowed with v = v1, ,

a(u, v) =d

i=1

xiu xiv d x ,

l(v) =

f v d x .

It is easy to see that the bilinear form a(, ) (resp. the linear form l()) is continuous on H10 () H10 ()

(resp. on H10 ()) since we have

a(u, v) |u|1,|v|1, u1, v1,

l(v) f0,v0, f0, v1, ,

where | |1, denotes the semi-norm on H1():

|v|1, =

d

i=1

xiv20,

1/2.

Furthermore, since

is an open bounded subset ofR

d

, the bilinear form a(, ) is H1

0 (

)-elliptic, as wehave:

v H10 () , a(v, v) 1

1 + C2v21, ,

thanks to Poincares inequality. According to Lax-Milgram theorem, there exists a unique function u H10 () such that

v H10 () ,d

i=1

xiuxiv dx =

f v d x ,

that is precisely the form (4.4).

Remark 4.2 We could have considered the Dirichlet boundary-value problem in the following formula-tion: given f H1(), find u such that

u = f in and u = 0 on .

As the right-hand side is not smooth, we will not look for a classical solution (u C2()) for this problem.Since u is in H1(), it seems reasonable to look for u H1() and thus in H10 () because of theboundary condition. Let consider v H10 (). The density of D() in H

10 () leads to write:

v H10 () , w H1() , w, v = w, vL2() .


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Hence, if u is a solution of the problem in H1(), then u is also a solution of the weak formulation: findu H10 () such that

v H10 () , u, vL2() = f, v .

Conversely, suppose u is a solution of the problem above and consider v = D():

D() , f, v = u, L2() = u, ,

hence, in the distributional sense:u = f ,

and since f H1(), this equality is posed in H1(). To solve this problem, we have to show thatthe weak formulation has a unique solution. This can be achieved by invoking Lax-Milgram theorem withV = H10 (), V

= H1(), a(u, v) = u, vL2() and l(v) = f, v. H10 () is a Hilbert space endowed

with the norm | |1, (equivalent to the norm 1, thanks to Poincares inequality) and u, vL2() isthe inner product associated with this norm. Hence, the hypothesis of Lax-Milgram theorem are satisfiedfor M = 1 and = 1, the problem has a unique solution and moreover,

|u|1,

f1,

. (4.9)If the function f has more regularity, the solution u will also have more regularity. For instance, iff L2(), then u H2() since u L2() and all the second derivatives are in L2().

Definition 4.3 An open set is said to be Lipschitz (or Lipschitz continuous) if it is bounded and if inthe vicinity of any point of its boundary, the boundary can be locally parametrized by a Lipschitz function, the domain being located on one side of the boundary.An open set is said to be of class Cm,1 if the Lipschitz function can be replaced by a Cm-continuousfunction, whose derivatives of order m are Lipschitz continuous.

Theorem 4.3 1. Suppose is of classC1,1 or is a convex polygon (polyhedron). Then, the operator

is an isomorphism of H2

(

) H1

0 (

) on L2

(

).2. Suppose is an arbitrary polygon inR2. For each t such that 1 < t 4/3, the operator is an

isomorphism of W2,t() H10 () on Lt().

3. Suppose is a Lipschitz polyhedron inR3. The operator is an isomorphism ofH3/2()H10 ()on L3/2().

Then, under the Lax-Milgram hypothesis, there exists a constant C such that

u H2() H10 () , u2, Cu0, .

Remark 4.3 This result is not true if is a polygon with a concave corner, i.e. there exists right-handside terms f for which the solution of the Dirichlet problem is not in H2().

4.3.2 Non-homogeneous Dirichlet conditions

Now, we consider the non-homogeneous Dirichlet problem:Given f H1() and given g H1/2(), find u H1() such that:

u = f in

u = g on .(4.10)


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Notice that this problem has at most one solution. The aim is to retrieve a homogeneous problem thatwe can solve. The trace theory indicates that if is C1-continuous and if g H1/2() then thereexists a function ug H

1() such that 0(ug) = g. In other words, the trace of ug on is the functiong. This function ug is called a Dirichlet lift ofg in . We pose u0 = u ug and our problem is equivalentto the homogeneous problem:

Given f H1

() and g H1/2

(), find u0 H1

() such that

u0 = f + ug in

u0 = 0 on .(4.11)

Notice first that ifug H1() then ug D

() but ug / L2(). However, since ug H

1(), thisproblem has a unique solution u0. Hence, u = u0 + ug is solution of the initial problem (4.10) that has atmost one solution, then this solution is unique and is independent of the Dirichlet lift function ug. Since,

ug1, ug0, = |ug|1, ,

the inequality (4.9) leads to write:

|u0|1, f1, + ug1, f1, + |ug|1, .

Hence, for every lift ug of the function g,

u1, u01, + ug1, (C2 + 1)1/2f1, + ((C

2 + 1)1/2 + 1)ug1, ,

and then

u1, (C2 + 1)1/2f1, + ((C

2 + 1)1/2 + 1) inf0v=g

v1,

(C2 + 1)1/2f1, + ((C2 + 1)1/2 + 1)gH1/2() .

Finally, we enounce several results that summarize and extend the previous remarks.

Theorem 4.4 (interior regularity) Let consider Rd an open bounded domain such that theDirichlet problem is well-posed. We note u H10 () the weak solution. Then, if f H

mloc() for a

certain m 0, then u Hm+2loc (), where

Hmloc() = {u D() , u Hm(Rd) , D()} .

Theorem 4.5 (global regularity) Let u be the weak solution of the Dirichlet problem (??). If is bounded and Cm+2-continuous (or if = Rd+) and if f H

m() (m 0), then u Hm+2().Furthermore, there exsts a constant Cm > 0 (independent of u and f) such that:

uHm+2() Cm fHm() .

Theorem 4.6 If isC1-continuous and bounded in one direction, the non-homogeneous Dirichlet prob-lem (4.10) admits:


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1. A unique weak solution u H1() if f L2() and g H1/2(). The solution u is such that:

uH1() C(fL2() + gH1/2()) ,

where the constant C is independent from u, f and g.

2. interior regularity: if f Hmloc() then u Hm+2loc () for every m N.

3. global regularity: if is bounded and of class Cm+2 or if = Rd+, f Hm() and g

Hm+3/2() (m N) then u Hm+2(). Furthermore, we have the following estimate:

uHm+2() C(fHm() + gHm+3/2()) ,

where the constant C is independent from u, f and g.

Proof. The trace theory indicates that if is C1-continuous and if g H1/2() then there exists a function

ug H1() such that 0(ug) = g. We consider the problem (4.11) as being posed in D(). Hence, we have:

u0,DD = f, DD + ug,DD , D() .

By definition of the derivation of distributions, we have:

di=1

xiu0, xiDD = f, DD di=1

xiug, xiDD , D() .

All the terms being L2() functions, we can rewrite the previous identity as:

u0 dx = f v d x ug dx , D() ,and the weak formulation of problem (4.11) becomes:

Find u0 H10 () such that:

u0 w dx =

f v d x

ug w d x , w H10 () .

In order to involve the Lax-Milgram theorem, we define the bilinear form a(v, w) =

u0 w dx and the linearform l(w) =

fw x

ug w dx. The linear form l is continuous since we have the estimate:

|l(w)| fL2()wL2() + ugL2()wL2() .

If we pose C = max(fL2(), ugL2()), we obtain:

|l(w)| CwH1() , w H10 () .

The Lax-Milgram theorem states that there exists a constant C > 0 such that:

u0H1() C(fL2() + ugL2()) C(fL2() + ugH1()) C(fL2() + gH1/2()) .

Since u = u0 + ug, it is easy to obtain the same type of estimate for the function u H1(). The theorem 4.4

gives the second item. Finally, if is Cm+2-continuous and if g Hm+3/2(), we can chose ug Hm+2() such

that ugHm+2() 1/2gHm+3/2(). Then, ug L2() and theorem 4.5 can be invoked to conclude.


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4.3.3 The Necas theorem

Now, we introduce an important theoretical result due to the Czech mathematician Necas (1929-).

Theorem 4.7 (Necas) LetV and W be two Hilbert spaces. We consider the abstract problem

Given l V, find u W such that : a(u, v) = l(v) , v V,where a is a bilinear continuous form on W V and f is a linear form on V. This problem is well-posedif and only if:

(c1) there exists a constant > 0 such that

infwW

supvV

a(w, v)

wWvV > 0 ,

(c2) we have:

v V , (w W , a(w, v) = 0) (v = 0) .

Under these assumptions, we have the stability estimate:

l V , uW 1

lV .

Proof. We consider the operator A L(W, V), u W Au V, defined by Au, vV,V = a(u, v) for v V.The problem (P) can be written as: find u W such that Au = l in V. Again, to ensure the existence anduniqueness of the solution, it is sufficient to show that A is a one-to-one mapping. This can be established usingthe same approach than for the Lax-Milgram theorem. The stability inequality results from :

uW supvV

a(u, v)vV

= supvV

l(v)vV

= lV .

Remark 4.4 The condition (c1) can be written in the following form:

w W, wW supvV

a(w, v)

vV.

This condition is equivalent to: Ker(A) = {0} and Im(A) = R(A) is closed.

The condition (c2) is often found as follows:

v V , (v = 0)

sup

wWa(w, v) > 0

.

This condition is equivalent to: Ker(At) = R(A) = {0}.

Both conditions (c1) and (c2) are optimal as they provide necessary and sufficient conditions forthe problem (P) to be well-posed.


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4.3.4 Saddle-point problems

We turn to a different type of problem, called a mixed problem. Let be a bounded open subset ofRd,the Stokes problem reads: find u, p in appropriate Hilbert spaces such that:

u + p = f in ,

div u = g in ,

u = 0 on .

(4.12)

where f : Rd and g : R are given functions and the unknowns u and p represent the velocityand the pressure in an (possibly incompressible) slow viscous flow confined in . The solution (u, p) isa saddle point as will be explained hereafter. Notice that we have a constraint of zero velocity on theboundary:

g = 0 since

u =

u = 0, using the divergence theorem.

Suppose the solution u, p is sufficienly smooth, we can take a test function v sufficiently smooth withvalues in Rd. Since the velocity vanishes on the boundary, we will consider a test function that vanisheson the boundary, just as we did for the homogeneous Dirichlet problem. We multiply each term of the

first equation by v and we integrate by parts on

. We obtain successively the terms:

v u =d

i=1

viui =d

i=1

ui vi

=d

i,j=1

xjui xjvi =

u : v ,

the term

v p =

p div v and finally, the following equation:

u : v p div v = f v . (4.13)In this equation, suppose f L2()d, then all terms are meaningful if we consider functions u H10 ()

d

and p L2(). To test the second equation, we consider a sufficiently smooth test function q:

q div u =

gq .

Since u H10 ()d, the left-hand side term is well defined if q L2(). However, since we notice that

div u =

g = 0, It is not necessary to test this equation by the constants. Hence, we can restrict thetest function space to the subspace of L2() defined as:

L2

0

() = {q L2() ,

q = 0} .

Moreover, if p is a solution, p + c is also a solution (where c is a constant). To overcome this problem,we add a condition of zero mean pressure. The weak form of this problem reads:

Given f L2()d and g L20(), find (u, p) H10 ()

d L20() such that:

u : v

p div v =

f v , v H10 ()d ,

q div u =

g q , q L20() .

(4.14)


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Let V and W be two Hilbert spaces. We introduce a specific form of the problem:

Given l V, find u W such that: a(u, v) = l(v), v Vwhich leads under certain assumptions to a saddle-point problem. Let consider f V, g W and twobilinear forms a, defined on V V, and b defined on W V. The new abstract problem reads:

Given f V, g W, find u V and p W such that:

(S)

a(u, v) + b(v, p) = f(v) , v V ,

b(u, q) = g(q) , q W .(4.15)

Remark 4.5 It is easy to check that in the Stokes problem we considered, V = H10 ()d, W = L20(),

a(u, v) =

u : v, b(v, p) =

p div v, f(v) =

f v and g(q) =

g q.

The following result can be considered as the natural extension of the Necas theorem for the abstractproblem (S).

Theorem 4.8 LetV andW be two Hilbert spaces, f V

, g W

, a L(VV,R

) andb L(VW,R

).The abstract problem (S) is well-posed if and only if:

> 0 , infuKer(b)

supvKer(b)

a(u, v)

uV vV ,

v Ker(b) , (u Ker(b), a(u, v) = 0) (v = 0) ,

(4.16)

where Ker(b) = {v V , q W, b(v, q) = 0} and

> 0 , infqW

supvV

b(v, q)

vV qW . (4.17)

Under these assumptions, we have the following estimates:

uV c1 fV + c2 gW ,

pW c3 fV + c4 gW .(4.18)

where the constants are given as: c1 =1 , c2 =

1(1 +

a ), c3 =

1(1 +

a ) and c4 =

a2

(1 + a ).

Proof. We introduce the operators A : X X, Au,vX,X = a(u, v) and B : X M (resp. Bt : M X),Bv,qM,M = b(v, q). Then, the problem (4.15) is equivalent to the problem of

finding u X and p M such that:

Au + Btp = f ,Bu = g ,

and this problem is well-posed if and only if the two following conditions are satisfied (admitted here):

(i) Ker(B) Ker(B) is an isomorphism,

(ii) B : X M is surjective.

According to this result, we admit here that the problem is well-posed. There exists ug X such that Bug = gand ugX gM . Introducing the variable = u ug leads to write:

v Ker(B) , a(, v) = f(v) a(ug, v) .


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By observing that:|f(v) a(ug, v)| (fX + a ugX) vX

(fX +a

gM) vX ,

and by taking the supremum on v Ker(B), we deduce that:

X fX +a

gM .

Using the triangle inequality uX u ugX + ugX leads to the first inequality. The stability inequality onp results from pM B

tpX, thus

pM a uX + fX .

The results follows using the previous estimate on uX .

Remark 4.6 If the bilinear form a(, ) is V-elliptic then the conditions (4.16) are satisfied.

We noticed before that if the bilinear form a(, ) is symmetric and positive, such abstract problem can

be interpreted as a minimization problem. Here, if the bilinear form a(, ) is symmetric positive, thisminimization problem is transformed into a saddle-point problem.

Definition 4.4 Given a linear mapping L : V W R, we say that (u, p) is a saddle-point of L if:

(v, q) V W , L(u, q) L(u, p) L(v, p) .

Proposition 4.1 (u, p) is a saddle-point of the linear mapping L if and only if:

supqW

L(u, q) = infvV

supqW

L(v, q) = L(u, p) = supqW

infvV

L(v, q) = infvV

L(v, p) . (4.19)

Proof. According to the previous definition, we have:

infvV

supqW

L(v, q) supqW

L(u, q) L(u, p) infvV

L(v, p) supqW

infvV

L(v, q) .

Moreover, for every (v, q) V W, we have

infvV

L(v, q) L(v, q) supqW

L(v, q) ,

and we deduce thensupqW

infvV

L(v, q) infvV

supqW

L(v, q) .

In other words, we deduce

infvV supqWL(v, q) = supqWL(u, q) = L(u, p) = infvV L(v, p) = supqW infvV L(v, q) .

and the results follows.

Theorem 4.9 Suppose the bilinear form a(, ) is symmetric and positive. The pair(u, p) is solution ofthe problem (S) if and only if (u, p) is a saddle-point of the functional

L(v, q) =1

2a(v, v) + b(v, q) f(v) g(q) . (4.20)


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Proof. Suppose (u, p) V W. On the one hand, we have

(q W, L(u, q) L(u, p)) (q W, b(u, qp) g(qp))

(q W, b(u, q) = g(q)) ,

On the other hand, we introduce the functional J(v) = 12a(v, v) + b(v, p) f(v).

(v V, L(u, p) L(v, p)) (J(u) = minvV

J(v))

(v V, a(u, v) + b(v, p) = f(v)) ,

From these inequalities we deduce that both equalities of the problem (S) are satisfied.

Corollary 4.1 Suppose the bilinear form a(, ) is symmetric and positive and the conditions (4.16) and(4.17) are satisfied. Then,

(i) The problem (S) has a unique solution that is the unique saddle-point of the functional L,

(ii) This solution satisfies the equalities (4.19).

The next step in the analysis of the weak formulation of eliptic problems is related to the approxi-

mation of the solution of such problems.

4.4 Variational approximation of elliptic problems

This section is devoted to the approximation of an abstract problem using a numerical method.

4.4.1 Abstract theory

We return to the general situation for the elliptic problems and we suppose given

(i) a Hilbert space V on R, endowed with a norm denoted ,

(ii) a bilinear form a(, ) continuous on V V and V-elliptic, i.e. such that:u V , v V , a(u, v) Mu v

v V , a(v, v) v2 .

(iii) a linear form l() continuous on V.

And, according to Lax-Milgram theorem, we know that the problem:(P) Find u V such that v V , a(u, v) = l(v),

has a unique solution. In this context, the Hilbert space V is of infinite dimension; In order to obtaina numerical approximation of the solution u, we will consider replacing the problem (P) by a discreteproblem posed in a finite dimensional space, denoted Vh, where h represents a discretization param-

eter intended to tend towards 0. For the sake of simplicity, we will only consider here a conformingapproximation for which Vh V.Under these assumptions, the bilinear form a(, ) and the linear form l() are defined on Vh Vh and

Vh, respectively and we consider the discrete problem:Find u V such that:

(Ph) vh Vh , a(uh, vh) = l(vh) . (4.21)

In this case, the discrete problem is called a Galerkin approximate problem, named after the Russianmathematician B.G. Galerkin (1871-1945). More generally, this approach is called a Galerkin method.

Under the three hypothesis (i) (iii) given hereabove, we have the following result.


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4.4. Variational approximation of elliptic problems 67

Theorem 4.10 (Lax-Milgram) The discrete problem (Ph) defined by (4.21) has a unique solution uhin Vh and we have the estimate:

l V , uhV 1

lV .

Proof. This result can be obtained by a direct application of the Lax-Milgram theorem , since Vh is a Hilbertspace for the norm associated with V.

Now, we analyze the error commited by replacing the solution u of he problem (P) by the solution uhof the problem (Ph). The following result, due to the French mathematician Jean Cea (1933-), helps tounderstand why the variational approximation method is so interesting. It states that the discretizationerror of the problem (P) is of the same order than the error commited by approximating V by Vh.

Lemma 4.2 (Cea) Under the hypothesis of the Lax-Milgram theorem, we have the error estimate:

u uh M

inf

vhVh

u vh . (4.22)

Proof. Suppose vh Vh. We pose wh = vh uh. Since Vh is a subspace ofV, wh Vh and thus wh V. Hence,we have

a(u uh, wh) = 0 ,

ora(u uh, u uh) = a(u uh, u vh) .

From the hypothesis (i) (iii), we deduce that

u uh2 a(u uh, u uh) = a(u uh, u vh) Mu uh u vh ,

and the results follows.

Remark 4.7 When the bilinear form a(, ) is symmetric, we have, for all vh Vh:

a(u vh, u vh) = a(u uh, u uh) + a(uh vh, uh vh)

and thusa(u uh, u uh) = inf

vhVha(u vh, u vh) .

and then, the constant in the error estimate is replaced by

M .

The theorem shows that the evaluation of the error is equivalent to the evaluation of the quantityinfvhVh u vh, in other words, it consists in evaluating the distance in V between the solution u of

the problem (P) and the subspace Vh of V.

Theorem 4.11 Suppose that there exists a subspaceV V, dense inV and a linear mappingrh : V Vhsuch that:

v V, limh0

v rh(v) = 0 .

Then, the approximation method converges, i.e.

limh0

u uh = 0 .


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Proof. Let > 0, since V is dense in V, there exists an element v V such that

u v

2C, C =

M

.

For such element v, there exists h() > 0 such that

h h() , u rh(v) 2C .

For a sufficiently small h h(), we have

u uh C(u v + v rh(v)) .

and the result follows.

When the variational approximation converges, it is interesting to study its rate of convergence, whenh 0. In general, we will look for a bound on the approximation error of the form

u uh C(u)hk , k > 0 ,

where C(u) is a constant independent of h and k > 0 is a constant. In such a case, the approximation is

said to be of order k and we write:u uh = O(h

k) .

The approximation spaces Vh of V must be carefully chosen, and several criteria can be used todetermine the appropriate spaces.

(i) Vh must have a basis (1, . . . ,N) such that the coefficients a(i,j ) and l(i) are easy to computenumerically, and such that the resulting linear system is not too difficult to solve with numericalalgorithms,

(ii) Vh must be chosen so that the approximation method converges and so that the numerical solutionis an accurate approximation of the exact solution u.

Unfortunately, the requirements are often contradictory and a good compromise must be sought.

4.4.2 Linear system

The problem (Ph) is a linear system to solve. If N = dim Vh, suppose (1, . . . ,N) is a basis of Vh. Thestiffness matrix A RN,N of the system is the matrix of coefficients:

aij = a(j,i) , 1 i,j, N .

The solution uh can be decomposed in the basis of Vh as:

uh =

N

i=1

uii ,

with the vector U = (ui)1iN RN. Denoting the right-hand side term of equation (4.21) F = (fi)

RN, where

fi = l(i) 1 i N ,

leads us to write the problem (Ph) in the matrix form as:

AU = F .

In the discrete forms of the Lax-Milgram and Necas theorems, we face the following conditions:


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4.4. Variational approximation of elliptic problems 69

(c0h) (Lax-Milgram) there exists a constant h > 0 such that

uh Vh , a(uh, vh) huh2V .

(c1h) (Necas) there exists a constant h > 0 such that

infwhWh

supvhVh

a(wh, vh)whW vhV

h ,

(c2h) (Necas) we have:vh Vh , ( sup

whWh

a(wh, vh) = 0) (vh = 0) .

Proposition 4.2 We have the following properties:

1. (c0h) A is positive definite,

2. (c1h) Ker(A) = {0},

3. (c2h) rank A = dim Vh.

Proof.

1. Suppose (c0h) holds. A direct calculation gives

X RN , AX,XN = a(,) 2V cX

2N ,

where , N denotes the Euclidean scalar product in RN, N the associated norm and for X R

N,

X = (Xi)1iN, =N

i=1 Xii Vh. This result shows that A is positive definite. Conversely, if thematrix A is positive definite, for every X RN, with XN = 1, we have AX,XN > 0. hence, we have

Vh , = 0 , a(,) = AX,XN =

A

X

XN,

X

XN

N

X2N

cX2N c2V ,

noticing that the unit sphere ofRN is compact and thus the continuous mapping X AX,XN is boundedby below by a constant c > 0.

2. We have the following equivalence:

X Ker(A) i ,Ni=1

aijXj = 0 i , a(,i) = 0

supvVh

a(, v) = 0 ,

we deduce

(c1h) ( Vh , ( supvVh

a(, v) = 0) ( = 0))

Ker(A) = {0} .

Conversely, suppose Ker(A) = {0}. We consider a sequence (wn)nN Vh with wnW = 1 and such that

supvhVh

a(wn, v)

vV

1

n.

From the compacity of the unit sphere of Vh, we deduce that any subsequence also denoted (wn) tendstowards w when n . The limit is such that wV = 1 and supvVh a(w, v) = 0. This means that

W Ker(A), where w =N

i=1 Wii. Hence, W = 0 which is in contradiction with the assumption thatwV = 1.


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3. proceed similarly with At.

Remark 4.8 If the bilinear form a(, ) is symmetric so is the matrix A.

4.4.3 Approximation of a saddle-point problem

We consider here the approximation of the abstract problem (S). Suppose Vh is a subspace of V, Wh isa subspace of W, both of finite dimensions and we consider the discrete problem:

(Sh)

find uh Vh and ph Wh such that

a(uh, vh) + b(vh, ph) = f(vh) , vh Vh ,

b(uh, qh) = g(qh) , qh Wh .

(4.23)

The following result establishes the existence and uniqueness conditions for a solution to the discreteproblem.

Theorem 4.12 The discrete problem (Sh) is well-posed if and only if two conditions are satisfied:

h > 0 , infuhKer(Bh)

supvhKer(Bh)

a(uh, vh)

uhV vhV h , (4.24)

where Bh : Vh W

h is the operator induced by the bilinear form b on the discrete spaces: Bhvh, qh =b(vh, qh) and Ker(Bh) denotes the kernel of Bh:

Ker(Bh) = {vh Vh , b(vh, qh) = 0 , qh Wh} ,

and if

h > 0 , infqhWh supvhVh

b(vh, qh)

vhV qhW h . (4.25)

and the solution (uh, ph) verifies the following estimates:

u uhV c1h infvhVh

u vhV + c2h infqhWh

p qhW , (4.26)

with c1h =

1 + ah

1 + bh

, c2h =

bh

if Ker(Bh) Ker(B) and c2h = 0 if Ker(Bh) Ker(B) and

p phW c3h infvhVh

u vhV + c4h infqhWh

p qhW , (4.27)

with c3h

= c1h

a

hand c

4h= 1 + b

h+ c

2h

a

h.

Proof. It consists in using Theorem 4.8 while noticing that in finite dimension, the first condition on h impliesthe first two conditions o the theorem (the injectivity and surjectivity are equivalent notions for an endomorphismin finite dimension).We introduce the notation

Zh(g) = {wh Xh , b(wh, qh) = g(qh) , qh Mh} .

Let consider vh Xh. Since Bh satisfies (4.25), there exists rh Xh such that

qh Mh , b(rh, qh) = b(u vh, qh) and hrhX bu vhX .


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4.5. Application to boundary-value problems in R 71

Since b(rh + vh, qh) = g(qh), then rh + vh belongs to the space Zh(g). We pose wh = rh + vh and we notice thatuh wh Ker(Bh) and we have:

huh whx supyhKer(Bh)

a(uh wh, yh)

yhX

supyhKer(Bh)

a(uh u, yh) + a(u wh, yh)

yhX

supyhKer(Bh)

b(yh, p ph) + a(u wh, yh)

yhX

If Ker(Bh) Ker(B), then b(yh, p ph) = 0 for yh Ker(Bh) and thus

huh whX a u whX .

The triangle inequality yields

u uhX

1 +

a

h

u whX .

In the general case, we have b(yh, ph) = b(yh, qh) = 0 for every qh Mh since yh Ker(Bh). Hence

huh whX a u whX + b p qhM .

The triangle inequality yields

u uhX

1 +

a

h

u whX +

b

hp qhM .

By noticing that the inequality u whX u vhX + rhX can be written as follows:

u whX

1 +

b

h

u vhX ,

we obtain the desired result.

Since for every vh Xh, b(vh, p ph) = a(uh u, vh), we obtain by considering qh Mh:

vh Xh , b(vh, qh ph) = a(uh u, vh) + b(vh, qh p) .

And by taking into account the condition (4.25), we deduce:

hqh phM a u uhX + b p qhM .

The final result is otained using the triangle inequality.

4.5 Application to boundary-value problems in R

We will briefly present the weak formulation and the variational approximation concepts on two basicboundary-value problems in one dimension.

4.5.1 Dirichlet problem

Let consider the homogeneous Dirichlet boundary-value problem in one dimension:

d

dx

du

dx

+ u = f , x =]0, 1[ ,

u(0) = u(1) = 0 .

, (4.28)


20/24


for and given functions of L() such that

(x) > 0 , (x) 0 , a.e. in .

We pose

a(u, v) = 1

0

(uv + uv) dx ,

and the weak formulation of this problem consists in finding u V = H10 () solving:

v H10 () , a(u, v) =

10

f v d x . (4.29)

The bilinear form a(, ) est continuous on V V and V-elliptic, thus, for any f L2(), the variationalproblem (4.29) is well-posed and has a unique solution u V.

To construct an approximation uh ofu, we consider the subspace Vh V composed of piecewise affinecontinuous functions on intervals. Let introduce the step h = 1N+1 , for N N

given, and the pointsxn = nh, 0 n N + 1. The latter subdivide the interval into N + 1 subintervals Kn = [xn, xn+1] oflength h. The subspace Vh is chosen a:

Vh = {v V , v|Kn P1 , 0 n N} ,where P1 denotes the space of polynomial functions of degree lesser than or equal to one. A functionv V = H10 () is (almost everywhere equal to) a continuous function on [0, 1] and satisfies the conditionsv(0) = v(1) = 0. Furthermore, any continuous function on , piecewise C1 on intervals and such thatv(0) = v(1) = 0 is a function of H10 (). This leads to consider Vh as:

Vh = {v C0() , v(0) = v(1) = 0 , v|Kn P1 , 0 n N} .

The dimension of Vh is exactly N and the sequence of functions (n)1nN Vh defined by:

i(xj) = i,j , 1 j N ,

forms a basis of Vh. Each function i can be written as:

i = 1 |x xi|h x [xi1, xi+1] ,0 otherwise

Hence, a function uh Vh can be uniquely determined using its coordinates un = uh(xn) in the basis(n)1nN. The variational method consists in approaching the solution u V by the function uh =N

j=1 ujj Vh, where the (uj )1jN are solution of the linear system:

Nj=1

a(j ,i)uj =

10

fj dx , 1 i N .

Actually, the linear system is relatively simple since the support of a function j is the interval [xj1, xj+1].This leads to consider the system:

a(1,1)u1 + a(2,1)u2 =

10

f1 dx

...

a(i1,i)ui1 + a(i,i)ui + a(i+1,i)ui+1 =

10

fi dx

...

a(N1,N)uN1 + a(N,N)uN =

10

fN dx


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4.5. Application to boundary-value problems in R 73

The next step consists in evaluating the coefficients of this linear system and specifically all integrals.To this end, we introduce the notations:

i+1/2 =1

h

xi+1xi

(x) dx ,

i+1/2 =3

h3xi+1

xi(xi+1 x)2(x) dx ,

i+1/2 =6

h3

xi+1xi

(xi+1 x)(x xi)(x) dx ,

+i1/2 =3

h3

xi+1xi

(x xi+1)2(x) dx ,

and

fi+1/2 =2

h2

xi+1xi

(xi+1 x)f(x) dx ,

f+i+1/2 =2

h2

xi+1xi

(x xi+1)f(x) dx ,

yielding to the following linear system:

1

h(1/2 + 3/2) +

h

3(+1/2 +

3/2

u1 +

1

h3/2 +

h

63/2

u2 =

h

2(f+1/2 + f

3/2) ,

...

1

hi1/2 +

h

6i1/2

ui1 +

1

h(i1/2 + i+1/2) +

h

3(+i1/2 +

i+1/2)

ui+

1

hi+1/2 +

h

6i+1/2

ui+1 =

h

2(f+i1/2 + f

i+1/2)

...

1h

N1/2 +h

6

N1/2uN1 + 1h

(N1/2 + N+1/2) +h

3

(+N1/2 + N+1/2) uN

=h

2(f+N1/2 + f

N+1/2)

The last issue that remains to be addressed concerns the right-hand side term. If the function f is anarbitrary function, the quantities fi+1/2 and f

+i+1/2 cannot be evaluated directly, they must be evaluated

using a numerical integration scheme. Given a continuous function g defined on a bounded interval [a, b],the mid-point formula allows to compute (exactly if g is affine) the integral of g on [a, b] as:b

ag(x) dx (b a)g

a + b

2

.

We proceed similarly for evaluating the quantities involving the functions and . It is important tocarefully evaluate such quantities as it affects the stiffness matrix A of the system.

The choice of the basis (n)1nN as described above leads to a tridiagonal matrix A. Thus, efficientmethods exist to resolve the system numerically (cf. Chapter 5).

4.5.2 Neumann problem

Let consider the Neumann boundary-value problem in one dimension:

d

dx

du

dx

+ u = f , x =]0, 1[ ,

u(0) = u(1) = 0 .

, (4.30)


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for and given functions of L() such that

(x) > 0 , (x) 0 , a.e. in .

We have already seen that for any given f L2(), this problem is well-posed and the solution u H1().Considering the subdivision of into N + 1 subintervals Kn = [xn, xn+1] of length h, we introduce

the space Vh H1() defined as:

Vh = {v C0() , v|Kn P1 , 1 n N} .

It is a subspace ofH1() of dimension N+2. A basis of this space is given by the functions (n)0nN+1:

0(x) =

1

x

hif 1 x h

0 otherwise

i(x) =

1 |x xi|

hif xi1 x xi+1

0 otherwise

N+1(x) =

1

1 x

hif 1 h x 1 ,

0 otherwise

The variational approximation method consists in searching for the solution uh =N+1

j=0 ujj Vhsolving the linear system:

1

h(1/2 +

h

3(1/2

u0 +

1

h1/2 +

h

61/2

u1 =

h

2(f1/2) ,

...

1

hi1/2 +h

6 i1/2

ui1 +1

h (i1/2 + i+1/2) +h

3 (+i1/2 +

i+1/2)

ui+

1

hi+1/2 +

h

6i+1/2

ui+1 =

h

2(f+i1/2 + f

i+1/2)

...

1

hN+1/2 +

h

6N+1/2

uN +

1

hN+1/2 +

h

3+N+1/2

uN+1 =

h

2f+N+1/2

As for the Dirichlet boundary-value problem, the stiffness matrix A is tridiagonal and symmetric positivedefinite.

Remark 4.9 In the Neumann problem, the numerical solutionuh does not verify the boundary conditions

uh(0) = u

h(1) = 0, neither a priori nor a posteriori.

4.6 Maximum principles

We conclude this chapter by a few words on maximum principles.


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4.7. Exercises 75

4.7 Exercises

Exercise 4.1 Suppose Rd is an open set and f L2() is a given function. We consider thefollowing Neumann problem:

u + u = f , on

un

= 0 , on .(4.31)

1. Show that the weak formulation of this problem is:

v H1() ,

u v + uv dx =

f v d x ,

and that there is a unique solution to this problem in H1().

2. Sho wthat the solution u H1() of this problem verifies u + u = f in D().

3. Suppose is of class C2 and thus u H2(). In what sense u is a solution of the initial problem ?


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Weak Formulation