Chapter 4. Weak formulation of elliptic problems

Chapter 4 

Weak formulation of elliptic problems 

”Alan Turing is reported as saying that PDE’s are made by God, the 

boundary conditions by the Devil ! The situation has changed, Devil 

has 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 and 

physical partial differential models. The aim is to introduce the notion of weak formulation that give 

access to existence and uniqueness results for the solutions and that is well suited for the numerical 

approximation 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 Ω ⊂ R d with a piecewise C 1 continuous boundary ∂Ω and let denote by 

ν the unit outer normal vector to Ω. We first recall a Green’s identity. 

Lemma 4.1 For every function u ∈ H 2 (Ω) and every function v ∈ H 1 (Ω), we have the Green’s identity: 

� 

− (∆u)v dx = 

Ω 

d� 

� 

i=1 

∂xi 

Ω 

53 

� 

u ∂xiv dx − 

∂Ω 

∂u 

v dσ . (4.1) 

∂ν

54 Chapter 4. Weak formulation of elliptic problems 

We consider the following model problem: 

Given a real-valued function f ∈ L 2 (Ω), find a function u defined in Ω solving 

−∆u = f in Ω (4.2) 

u = 0 on ∂Ω . (4.3) 

Suppose the solution u is sufficiently smooth, for example u ∈ H2 (Ω). Then, we can multiply each side 

of the equation (4.2) by a test function v ∈ H1 0 (Ω) and integrate on Ω to obtain: 

� 

� 

− (∆u)v dx = fv dx . 

Thanks to Green’s identity (4.1) and by noticing that v|∂Ω = 0, we obtain: 

∀v ∈ H 1 0 (Ω) , 

Ω 

d� 

� 

i=1 

∂xi 

Ω 

Ω 

u ∂xi v dx = 

� 

Ω 

fv dx . (4.4) 

The specific boundary conditions and the Theorem 2.21 yield to conclude that the solution u of the 

problem (4.4) is such that: 

u ∈ H 1 0 (Ω) 

and the initial problem is now replaced by the following problem: 

Given a function f ∈ L2 (Ω), find a function u ∈ H1 0 (Ω) 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 ∈ H1 0 (Ω) is a solution of the problem (4.4) if and only if: 

∀ϕ ∈ D(Ω) , 

d� 

� 

i=1 

∂xi 

Ω 

u ∂xi ϕ dx = 

� 

Ω 

fϕ dx , 

since H1 0 (Ω) is the closure of D(Ω) in H1 (Ω) (cf. Definition 2.14). Consequently, if u satisfies the equation 

(4.4), then the initial equation −∆u = f is satisfied on Ω in the distributional sense and since f ∈ L2 (Ω), 

this equation is verified in L2 (Ω), thus almost everywhere in Ω. Finally, u ∈ H1 0 (Ω) leads to satisfy the 

boundary condition according to the trace theorem on the boundary ∂Ω. 

4.2 Neumann problem 

Let consider an open bounded domain Ω ⊂ R d with a piecewise C 1 continuous boundary ∂Ω. We now 

focus on the following problem: 

Given a function f ∈ L 2 (Ω), find a function u defined on Ω solving 

−∆u + u = f in Ω (4.5) 

∂u 

∂ν 

= 0 on ∂Ω . (4.6) 

Suppose the solution u ∈ H2 (Ω). Then, following the same procedure as for the Dirichlet problem, we 

obtain: 

∀v ∈ H 1 (Ω) , 

d� 

� 

� � 

∂xiu ∂xiv dx + uv dx = fv dx . 

Ω 

Ω 

Ω 

(4.7) 

i=1

4.3. Abstract variational problems 55 

The original boundary-value problem is then replaced by the following: 

Given a function f ∈ L 2 (Ω), find a solution u ∈ H 1 (Ω) verifying equation (4.7). 

Any sufficiently smooth solution u of the Neumann problem is solving the previous problem. Conversely, 

if u ∈ H 1 (Ω) solves the problem (4.7), we have 

∀ϕ ∈ D(Ω) , 

d� 

� 

i=1 

∂xi 

Ω 

� 

u ∂xiϕ dx + uϕ dx = 

Ω 

� 

Ω 

fϕ dx , 

and since u and f are L 2 (Ω) functions, the initial equation −∆u + u = f is satisfied in the distributional 

sense, in L 2 (Ω). When u ∈ H 2 (Ω), we can show, using the Green’s formula, that equation 4.7 is equivalent 

to the initial equation and that: 

∀v ∈ H 1 (Ω) , 

� 

∂Ω 

∂u 

v dσ = 0 , 

∂ν 

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

∀w ∈ L 2 (∂Ω) , 

� 

∂u 

w dσ = 0 , 

∂ν 

which leads to the Neumann boundary condition in the L 2 (Ω) 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 large 

class 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) ≤ M�u� �v� , 

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

endowed with the dual norm 

l(v) 

�l�V ′ = sup 

v∈V �v� 

v�=0 

. 

Then, we consider the general variational problem: 

� 

′ 

Given l ∈ V , find u ∈ V solving the problem 

(P) 

∀v ∈ V , a(u, v) = l(v) 

As such, this problem is certainly not well-posed. We will give a sufficient condition on the bilinear form 

a 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 problem 

and if the following linear stability property is satisfied: 

∂Ω 

∃C > 0 , ∀l ∈ V ′ , �u� ≤ C �l�V ′ . 

The objective is now to determine if and when a variational problem is well-posed. Two fundamental 

results will help us to solve this question. 

(4.8)


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 α > 0 such that 

∀v ∈ V , a(v, v) ≥ α�v� 2 . 

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 α, 

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

∀l ∈ V ′ , �u�V ≤ 1 

�l�V ′ , 

α 

(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 a 

unique 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 theorem 

again, 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 the 

problem (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 from 

V to V . A being continuous, the inverse A −1 is continuous from V to V . 

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

and thus, 

This shows that A is an injection, since: 

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

∀v ∈ V , α�v� 2 ≤ a(v, v) = 〈Av, v〉 ≤ �Av� �v� , 

∀v ∈ V , �Av� ≥ α�v� . 

v ∈ Ker(A) ⇔ Av = 0 ⇒ 〈Av, v〉 = 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)m∈N be a sequence of R(A) that converges 

to w in V . We have: 

�Avm − Avn� ≥ α�vm − vn� 

such 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: 

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

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

w = Av ∈ R(A) , 

α�v0� 2 ≤ 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.


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

and hence that A −1 is continuous. By noticing that: 

∀v ∈ V , �A −1 v� ≤ 1 

�v� , 

α 

〈wl, v〉 

�wl� = sup 

v∈V �v� 

v�=0 

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

l(v) 

= sup 

v∈V �v� 

v�=0 

= �l�∗ , 

�u� = �A −1 wl� ≤ 1 1 

�wl� = �l�V ′ , 

α α 

and the result follows. � 

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

minimization 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 

i.e. J(u) = inf 

v∈V J(v). 

∀v ∈ V , J(v) = 1 

a(v, v) − l(v) , 

2 

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 

a(v − u, v − u) 

2 

= J(u) + 1 

a(v − u, v − u) 

2 

≥ J(u) , 

which attests that u minimizes the function J on V . Conversely, let consider v ∈ V , λ ∈]0, 1] and we pose 

w = u + λv. A direct calculation gives: 

Dividing by λ and letting λ → 0 yields 

J(w) − J(u) = λ (a(u, v) − l(v)) + λ2 

a(v, v) ≥ 0 . 

2 

∀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 minimization 

of a quadratic functional on a Hilbert space V , which is the abstract formulation of 

numerous problems in calculus of variations. This explains why (P) is called a variational problem.


(ii) When the bilinear form a(·, ·) is symmetric and V -elliptic, the Lax-Milgram theorem indicates that 

the optimization problem infv∈V J(v) has a unique solution. The V -ellipticity of a can be seen as a 

property 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 of 

problem (P), if we pose 

V = H 1 0 (Ω) endowed with �v� = �v�1,Ω , 

a(u, v) = 

d� 

� 

i=1 

� 

l(v) = 

Ω 

∂xi 

Ω 

fv dx . 

u ∂xi v dx , 

It is easy to see that the bilinear form a(·, ·) (resp. the linear form l(·)) is continuous on H1 0 (Ω) × H1 0 (Ω) 

(resp. on H1 0 (Ω)) since we have 

where | · |1,Ω denotes the semi-norm on H 1 (Ω): 

a(u, v) ≤ |u|1,Ω|v|1,Ω ≤ �u�1,Ω �v�1,Ω 

l(v) ≤ �f�0,Ω�v�0,Ω ≤ �f�0,Ω �v�1,Ω , 

|v|1,Ω = 

� d� 

i=1 

�∂xi v�2 

0,Ω 

Furthermore, since Ω is an open bounded subset of Rd , the bilinear form a(·, ·) is H1 0 (Ω)-elliptic, as we 

have: 

∀v ∈ H 1 0 (Ω) , a(v, v) ≥ 

�1/2 

. 

1 

1 + C 2 �v�2 1,Ω , 

thanks to Poincaré’s inequality. According to Lax-Milgram theorem, there exists a unique function u ∈ 

H1 0 (Ω) such that 

d� 

� 

� 

u ∂xiv dx = fv dx , 

that is precisely the form (4.4). 

∀v ∈ H 1 0 (Ω) , 

i=1 

∂xi 

Ω 

Remark 4.2 We could have considered the Dirichlet boundary-value problem in the following formulation: 

given f ∈ H −1 (Ω), find u such that 

−∆u = f in Ω and u = 0 on ∂Ω . 

As the right-hand side is not smooth, we will not look for a classical solution (u ∈ C2 (Ω)) for this problem. 

Since ∆u is in H−1 (Ω), it seems reasonable to look for u ∈ H1 (Ω) and thus in H1 0 (Ω) because of the 

boundary condition. Let consider v ∈ H1 0 (Ω). The density of D(Ω) in H1 0 (Ω) leads to write: 

∀v ∈ H 1 0 (Ω) , ∀w ∈ H 1 (Ω) , 〈−∆w, v〉 = 〈∇w, ∇v〉 L 2 (Ω) . 

Ω


Hence, if u is a solution of the problem in H1 (Ω), then u is also a solution of the weak formulation: find 

u ∈ H1 0 (Ω) such that 

∀v ∈ H 1 0 (Ω) , 〈∇u, ∇v〉 L2 (Ω) = 〈f, v〉 . 

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

hence, in the distributional sense: 

∀ϕ ∈ D(Ω) , 〈f, v〉 = 〈∇u, ∇ϕ〉 L 2 (Ω) = −〈∆u, ϕ〉 , 

−∆u = f , 

and since f ∈ H −1 (Ω), this equality is posed in H −1 (Ω). To solve this problem, we have to show that 

the weak formulation has a unique solution. This can be achieved by invoking Lax-Milgram theorem with 

V = H1 0 (Ω), V ′ = H−1 (Ω), a(u, v) = 〈∇u, ∇v〉 L2 (Ω) and l(v) = 〈f, v〉. H1 0 (Ω) is a Hilbert space endowed 

with the norm |·|1,Ω (equivalent to the norm �·�1,Ω thanks to Poincaré’s inequality) and 〈∇u, ∇v〉 L2 (Ω) is 

the inner product associated with this norm. Hence, the hypothesis of Lax-Milgram theorem are satisfied 

for M = 1 and α = 1, the problem has a unique solution and moreover, 

|u|1,Ω ≤ �f�−1,Ω . (4.9) 

If the function f has more regularity, the solution u will also have more regularity. For instance, if 

f ∈ L 2 (Ω), then u ∈ H 2 (Ω) since ∆u ∈ L 2 (Ω) and all the second derivatives are in L 2 (Ω). 

Definition 4.3 An open set Ω is said to be Lipschitz (or Lipschitz continuous) if it is bounded and if in 

the vicinity of any point of its boundary, the boundary can be locally parametrized by a Lipschitz function 

ϕ, the domain being located on one side of the boundary. 

An open set Ω is said to be of class C m,1 if the Lipschitz function ϕ can be replaced by a C m -continuous 

function, whose derivatives of order m are Lipschitz continuous. 

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

−∆ is an isomorphism of H 2 (Ω) ∩ H 1 0 (Ω) on L2 (Ω). 

2. Suppose Ω is an arbitrary polygon in R 2 . For each t such that 1 < t ≤ 4/3, the operator −∆ is an 

isomorphism of W 2,t (Ω) ∩ H 1 0 (Ω) on Lt (Ω). 

3. Suppose Ω is a Lipschitz polyhedron in R 3 . The operator −∆ is an isomorphism of H 3/2 (Ω)∩H 1 0 (Ω) 

on L 3/2 (Ω). 

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

∀u ∈ H 2 (Ω) ∩ H 1 0 (Ω) , �u�2,Ω ≤ C �∆u�0,Ω . 

Remark 4.3 This result is not true if Ω is a polygon with a concave corner, i.e. there exists right-hand 

side terms f for which the solution of the Dirichlet problem is not in H 2 (Ω). 

4.3.2 Non-homogeneous Dirichlet conditions 

Now, we consider the non-homogeneous Dirichlet problem: 

Given f ∈ H −1 (Ω) and given g ∈ H 1/2 (∂Ω), find u ∈ H 1 (Ω) such that: 

−∆u = f in Ω 

u = g on ∂Ω . 

(4.10)


Notice that this problem has at most one solution. The aim is to retrieve a homogeneous problem that 

we can solve. The trace theory indicates that if ∂Ω is C 1 -continuous and if g ∈ H 1/2 (∂Ω) then there 

exists a function ug ∈ H 1 (Ω) such that γ0(ug) = g. In other words, the trace of ug on ∂Ω is the function 

g. This function ug is called a Dirichlet lift of g in Ω. We pose u0 = u − ug and our problem is equivalent 

to the homogeneous problem: 

Given f ∈ H −1 (Ω) and g ∈ H 1/2 (∂Ω), find u0 ∈ H 1 (Ω) such that 

−∆u0 = f + ∆ug in Ω 

u0 = 0 on ∂Ω . 


Notice first that if ug ∈ H 1 (Ω) then ∆ug ∈ D ′ (Ω) but ∆ug /∈ L 2 (Ω). However, since ∆ug ∈ H −1 (Ω), this 

problem has a unique solution u0. Hence, u = u0 + ug is solution of the initial problem (4.10) that has at 

most one solution, then this solution is unique and is independent of the Dirichlet lift function ug. Since, 

the inequality (4.9) leads to write: 

Hence, for every lift ug of the function g, 

and then 

�∆ug�−1,Ω ≤ �∇ug�0,Ω = |ug|1,Ω , 

|u0|1,Ω ≤ �f�−1,Ω + �∆ug�−1,Ω ≤ �f�−1,Ω + |ug|1,Ω . 

�u�1,Ω ≤ �u0�1,Ω + �ug�1,Ω ≤ (C 2 + 1) 1/2 �f�−1,Ω + ((C 2 + 1) 1/2 + 1)�ug�1,Ω , 

�u�1,Ω ≤ (C 2 + 1) 1/2 �f�−1,Ω + ((C 2 + 1) 1/2 + 1) inf 

γ0v=g �v�1,Ω 

≤ (C 2 + 1) 1/2 �f�−1,Ω + ((C 2 + 1) 1/2 + 1)�g� H 1/2 (∂Ω) . 

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

Theorem 4.4 (interior regularity) Let consider Ω ⊂ Rd an open bounded domain such that the 

Dirichlet problem is well-posed. We note u ∈ H1 0 (Ω) the weak solution. Then, if f ∈ Hm certain m ≥ 0, then u ∈ H 

loc (Ω) for a 

m+2 

loc (Ω), where 

H m loc (Ω) = {u ∈ D′ (Ω) , ϕu ∈ H m (R d ) , ∀ϕ ∈ D(Ω)} . 

Theorem 4.5 (global regularity) Let u be the weak solution of the Dirichlet problem (??). If ∂Ω 

is bounded and C m+2 -continuous (or if Ω = R d +) and if f ∈ H m (Ω) (m ≥ 0), then u ∈ H m+2 (Ω). 

Furthermore, there exsts a constant Cm > 0 (independent of u and f) such that: 

�u� H m+2 (Ω) ≤ Cm �f� H m (Ω) . 

Theorem 4.6 If Ω is C 1 -continuous and bounded in one direction, the non-homogeneous Dirichlet problem 

(4.10) admits:


1. A unique weak solution u ∈ H 1 (Ω) if f ∈ L 2 (Ω) and g ∈ H 1/2 (∂Ω). The solution u is such that: 

�u� H 1 (Ω) ≤ C (�f� L 2 (Ω) + �g� H 1/2 (∂Ω) ) , 

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

2. interior regularity: if f ∈ Hm loc (Ω) then u ∈ Hm+2 

loc (Ω) for every m ∈ N. 

3. global regularity: if ∂Ω is bounded and of class C m+2 or if ∂Ω = R d +, f ∈ H m (Ω) and g ∈ 

H m+3/2 (∂Ω) (m ∈ N) then u ∈ H m+2 (Ω). Furthermore, we have the following estimate: 

�u� H m+2 (Ω) ≤ C (�f� H m (Ω) + �g� H m+3/2 (∂Ω) ) , 

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

Proof. The trace theory indicates that if ∂Ω is C 1 -continuous and if g ∈ H 1/2 (∂Ω) then there exists a function 

ug ∈ H 1 (Ω) such that γ0(ug) = g. We consider the problem (4.11) as being posed in D ′ (Ω). Hence, we have: 

〈−∆u0, ϕ〉D ′ ×D = 〈f, ϕ〉D ′ ×D + 〈∆ug, ϕ〉D ′ ×D , ∀ϕ ∈ D(Ω) . 

By definition of the derivation of distributions, we have: 

d� 

i=1 

〈∂xi u0, ∂xi ϕ〉D ′ ×D = 〈f, ϕ〉D ′ ×D − 

All the terms being L 2 (Ω) functions, we can rewrite the previous identity as: 

� 

∇u0 · ∇ϕ dx = 

Ω 

and the weak formulation of problem (4.11) becomes: 

Find u0 ∈ H 1 0 (Ω) such that: 

� 

∇u0 · ∇w dx = 

Ω 

� 

� 

Ω 

Ω 

d� 

i=1 

∂xi ug, ∂xi ϕ〉D ′ ×D , ∀ϕ ∈ D(Ω) . 

� 

fv dx − ∇ug · ∇ϕ dx , ∀ϕ ∈ D(Ω) , 

Ω 

fv dx − 

� 

∇ug · ∇w dx , 

Ω 

∀w ∈ H 1 0 (Ω) . 

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

Ω ∇u0 · ∇w dx and the linear 

form l(w) = � � 

fw x − Ω Ω ∇ug · ∇w dx. The linear form l is continuous since we have the estimate: 

If we pose C = max(�f� L 2 (Ω), �∇ug� L 2 (Ω)), we obtain: 

|l(w)| ≤ �f� L 2 (Ω)�w� L 2 (Ω) + �∇ug� L 2 (Ω)�∇w� L 2 (Ω) . 

|l(w)| ≤ C�w� H 1 (Ω) , ∀w ∈ H 1 0 (Ω) . 

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

�u0� H 1 (Ω) ≤ C(�f� L 2 (Ω) + �∇ug� L 2 (Ω)) ≤ C(�f� L 2 (Ω) + �ug� H 1 (Ω)) ≤ C(�f� L 2 (Ω) + �g� H 1/2 (Ω)) . 

Since u = u0 + ug, it is easy to obtain the same type of estimate for the function u ∈ H 1 (Ω). The theorem 4.4 

gives the second item. Finally, if Ω is C m+2 -continuous and if g ∈ H m+3/2 (∂Ω), we can chose ug ∈ H m+2 (Ω) such 

that �ug� H m+2 (Ω) ≤ 1/2�g� H m+3/2 (Ω). Then, ∆ug ∈ L 2 (Ω) and theorem 4.5 can be invoked to conclude. �


4.3.3 The Nečas theorem 

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

Theorem 4.7 (Nečas) Let V 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-posed 

if and only if: 

(c1) there exists a constant α > 0 such that 

(c2) we have: 

inf 

w∈W sup 

v∈V 

a(w, v) 

�w�W �v�V 

≥ α > 0 , 

∀v ∈ V , (∀w ∈ W , a(w, v) = 0) ⇒ (v = 0) . 

Under these assumptions, we have the stability estimate: 

∀l ∈ V ′ , �u�W ≤ 1 

�l�V ′ . 

α 

Proof. We consider the operator A ∈ L(W, V ′ ), u ∈ W ↦→ Au ∈ V ′ , defined by 〈Au, v〉V ′ ,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 and 

uniqueness of the solution, it is sufficient to show that A is a one-to-one mapping. This can be established using 

the same approach than for the Lax-Milgram theorem. The stability inequality results from : 

α�u�W ≤ sup 

v∈V 

a(u, v) 

�v�V 

= sup 

v∈V 

l(v) 

�v�V 

= �l�V ′ . 

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

∀w ∈ W, α�w�W ≤ sup 

v∈V 

a(w, v) 

. 

�v�V 

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) ⇒ 

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

� 

� 

sup a(w, v) > 0 . 

w∈W 

• Both conditions (c1) and (c2) are optimal as they provide necessary and sufficient conditions for 

the problem (P) to be well-posed. 

�


4.3.4 Saddle-point problems 

We turn to a different type of problem, called a mixed problem. Let Ω be a bounded open subset of Rd , 

the Stokes problem reads: find u, p in appropriate Hilbert spaces such that: 

⎧ 

⎪⎨ −∆u + ∇p = f in Ω , 

⎪⎩ 

div u = g 

u = 0 

in Ω , 

on ∂Ω . 


where f : Ω → Rd and g : Ω → R are given functions and the unknowns u and p represent the velocity 

and the pressure in an (possibly incompressible) slow viscous flow confined in Ω. The solution (u, p) is 

a saddle point as will be explained hereafter. Notice that we have a constraint of zero velocity on the 

boundary: � 

� � 

Ω g = 0 since Ω ∇ · u = ∂Ω u · ν = 0, using the divergence theorem. 

Suppose the solution u, p is sufficienly smooth, we can take a test function v sufficiently smooth with 

values in R d . Since the velocity vanishes on the boundary, we will consider a test function that vanishes 

on 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 Ω. We obtain successively the terms: 

the term � 

Ω 

v · ∇p = − � 

Ω 

� 

− v · ∆u = 

Ω 

= 

d� 

� 

− 

d� 

� 

vi∆ui = ∇ui · ∇vi 

Ω 

i=1 

Ω 

d� 

� 

∂xj 

Ω 

ui ∂xjvi � 

= ∇u : ∇v , 

Ω 

i=1 

i,j=1 

p div v and finally, the following equation: 

� 

Ω 

� 

� 

∇u : ∇v − p div v = f · v . (4.13) 

Ω 

Ω 

In this equation, suppose f ∈ L 2 (Ω) d , then all terms are meaningful if we consider functions u ∈ H 1 0 (Ω)d 

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

� 

� 

q div u = gq . 

Ω 

Since u ∈ H 1 0 (Ω)d , the left-hand side term is well defined if q ∈ L 2 (Ω). However, since we notice that 

� � 

Ω div u = Ω g = 0, It is not necessary to test this equation by the constants. Hence, we can restrict the 

test function space to the subspace of L2 (Ω) defined as: 

L 2 0(Ω) = {q ∈ L 2 � 

(Ω) , 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 (Ω) d and g ∈ L2 0 (Ω), find (u, p) ∈ H1 0 (Ω)d × L2 0 (Ω) such that: 

⎧ � 

� 

� 

⎪⎨ ∇u : ∇v − p div v = f · v , ∀v ∈ H 

Ω 

Ω 

Ω 

⎪⎩ 

1 0 (Ω) d , 

� 

� 

q div u = g q , ∀q ∈ L 2 (4.14) 

0(Ω) . 

Ω 

Ω 

Ω 

Ω


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 ∈ V 

which leads under certain assumptions to a saddle-point problem. Let consider f ∈ V ′ , g ∈ W ′ and two 

bilinear 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 . 


Remark 4.5 It is easy to check that in the Stokes problem we considered, V = H 1 0 (Ω)d , W = L 2 0 (Ω), 

� 

a(u, v) = 

Ω 

� 

∇u : ∇v, b(v, p) = − 

Ω 

� 

p div v, f(v) = 

Ω 

f · v and g(q) = − � 

The following result can be considered as the natural extension of the Nečas theorem for the abstract 

problem (S). 

Theorem 4.8 Let V and W be two Hilbert spaces, f ∈ V ′ , g ∈ W ′ , a ∈ L(V ×V, R) and b ∈ L(V ×W, R). 

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

∃α > 0 , inf 

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

sup 

u∈Ker(b) v∈Ker(b) 

a(u, v) 

�u�V �v�V 

≥ α , 

∀v ∈ Ker(b) , (∀u ∈ Ker(b), a(u, v) = 0) ⇒ (v = 0) , 

∃β > 0 , inf 

q∈W sup 

v∈V 

Under these assumptions, we have the following estimates: 

where the constants are given as: c1 = 1 

α , c2 = 1 

β 

b(v, q) 

�v�V �q�W 

�u�V ≤ c1 �f�V ′ + c2 �g�W ′ , 

�p�W ≤ c3 �f�V ′ + c4 �g�W ′ . 

(1 + �a� 

α ), c3 = 1 

β 

Ω 

g q. 


≥ β . (4.17) 

(1 + �a� 

α ) and c4 = �a� 

β 2 (1 + �a� 

α ). 


Proof. We introduce the operators A : X → X, 〈Au, v〉X ′ ,X = a(u, v) and B : X → M ′ (resp. B t : M → X ′ ), 

〈Bv, q〉M ′ ,M = b(v, q). Then, the problem (4.15) is equivalent to the problem of 

finding u ∈ X and p ∈ M ′ such that: 

� Au + B t p = 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 = g 

and β�ug�X ≤ �g�M ′. Introducing the variable φ = u − ug leads to write: 

∀v ∈ Ker(B) , a(φ, v) = f(v) − a(ug, v) .


By observing that: 

|f(v) − a(ug, v)| ≤ (�f�X ′ + �a� �ug�X) �v�X 

�a� 

≤ (�f�X ′ + 

β �g�M ′) �v�X , 

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

α�φ�X ≤ �f�X 

�a� 

′ + �g�M ′ . 

β 

Using the triangle inequality �u�X ≤ �u − ug�X + �ug�X leads to the first inequality. The stability inequality on 

p results from β�p�M ≤ �Btp�X ′, thus 

β�p�M ≤ �a� �u�X + �f�X ′ . 

The results follows using the previous estimate on �u�X. � 

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, this 

minimization 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: 

sup L(u, q) = inf 

q∈W 

v∈V sup L(v, q) = L(u, p) = sup inf L(v, q) = inf L(v, p) . (4.19) 

q∈W 

q∈W v∈V v∈V 

Proof. According to the previous definition, we have: 

inf 

v∈V sup L(v, q) ≤ sup L(u, q) ≤ L(u, p) ≤ inf L(v, p) ≤ sup 

q∈W 

q∈W 

v∈V q∈W 

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

and we deduce then 

In other words, we deduce 

inf 

v ′ ∈V L(v′ , q) ≤ L(v, q) ≤ sup 

q ′ L(v, q 

∈W 

′ ) , 

sup 

q∈W 

inf L(v, q) ≤ inf 

v∈V v∈V sup L(v, q) . 

q∈W 

inf 

v∈V sup L(v, q) = sup L(u, q) = L(u, p) = inf L(v, p) = sup 

q∈W 

q∈W 

v∈V q∈W 

inf L(v, q) . 

v∈V 

inf L(v, q) . 

v∈V 

and the results follows. � 

Theorem 4.9 Suppose the bilinear form a(·, ·) is symmetric and positive. The pair (u, p) is solution of 

the problem (S) if and only if (u, p) is a saddle-point of the functional 

L(v, q) = 1 

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

2


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

(∀q ∈ W, L(u, q) ≤ L(u, p)) ⇔ (∀q ∈ W, b(u, q − p) ≤ g(q − p)) 

⇔ (∀q ∈ W, b(u, q) = g(q)) , 

On the other hand, we introduce the functional J(v) = 1 

2a(v, v) + b(v, p) − f(v). 

(∀v ∈ V, L(u, p) ≤ L(v, p)) ⇔ (J(u) = min 

v∈V 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 approximation 

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: 

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

∀u ∈ V , ∀v ∈ V , a(u, v) ≤ M�u� �v� 

∀v ∈ V , a(v, v) ≥ α�v� 2 . 

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 obtain 

a numerical approximation of the solution u, we will consider replacing the problem (P) by a ”discrete 

problem” posed in a finite dimensional space, denoted Vh, where h represents a discretization parameter 

intended to tend towards 0. For the sake of simplicity, we will only consider here a conforming 

approximation 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 Russian 

mathematician 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.

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 uh 

in Vh and we have the estimate: 

∀l ∈ V ′ , �uh�V ≤ 1 

�l�V ′ . 

α 

Proof. This result can be obtained by a direct application of the Lax-Milgram theorem , since Vh is a Hilbert 

space for the norm associated with V . � 

Now, we analyze the error commited by replacing the solution u of he problem (P) by the solution uh 

of the problem (Ph). The following result, due to the French mathematician Jean Céa (1933-), helps to 

understand why the variational approximation method is so interesting. It states that the discretization 

error of the problem (P) is of the same order than the error commited by approximating V by Vh. 

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

�u − uh� ≤ M 

α inf �u − vh� . (4.22) 

vh∈Vh 

Proof. Suppose vh ∈ Vh. We pose wh = vh − uh. Since Vh is a subspace of V , wh ∈ Vh and thus wh ∈ V . Hence, 

we have 

a(u − uh, wh) = 0 , 

or 

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

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

α�u − uh� 2 ≤ a(u − uh, u − uh) = a(u − uh, u − vh) ≤ M�u − 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 thus 

a(u − uh, u − uh) = inf a(u − vh, u − vh) . 

vh∈Vh 

� 

M 

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

The theorem shows that the evaluation of the error is equivalent to the evaluation of the quantity 

infvh∈Vh �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 subspace V ⊂ V , dense in V and a linear mapping rh : V → Vh 

such that: 

∀v ∈ V , lim 

h→0 �v − rh(v)� = 0 . 

Then, the approximation method converges, i.e. 

lim 

h→0 �u − uh� = 0 .


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

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

For a sufficiently small h ≤ h(ε), we have 

�u − v� ≤ ε 

M 

, C = 

2C α . 

∀h ≤ h(ε) , �u − rh(v)� ≤ ε 

2C . 

�u − uh� ≤ C(�u − v� + �v − rh(v)�) ≤ ε . 

and the result follows. � 

When the variational approximation converges, it is interesting to study its rate of convergence, when 

h → 0. In general, we will look for a bound on the approximation error of the form 

�u − uh� ≤ C(u)h k , 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 to 

determine the appropriate spaces. 

(i) Vh must have a basis (ϕ1, . . . , ϕN) such that the coefficients a(ϕi, ϕj) and l(ϕi) are easy to compute 

numerically, and such that the resulting linear system is not too difficult to solve with numerical 

algorithms, 

(ii) Vh must be chosen so that the approximation method converges and so that the numerical solution 

is 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 (ϕ1, . . . , ϕN) is a basis of Vh. The 

stiffness matrix A ∈ R N,N of the system is the matrix of coefficients: 

aij = a(ϕj, ϕi) , 1 ≤ i, j, ≤ N . 

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

uh = 

N� 

i=1 

uiϕi , 

with the vector U = (ui)1≤i≤N ∈ R N . Denoting the right-hand side term of equation (4.21) F = (fi) ∈ 

R N , where 

fi = l(ϕi) 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 Nečas theorems, we face the following conditions:

4.4. Variational approximation of elliptic problems 69 

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

(c1h) (Nečas) there exists a constant αh > 0 such that 

(c2h) (Nečas) we have: 

∀uh ∈ Vh , a(uh, vh) ≥ αh�uh� 2 V . 

inf 

sup 

wh∈Wh vh∈Vh 

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 

a(wh, vh) 

�wh�W �vh�V 

≥ αh , 

∀vh ∈ Vh , ( sup a(wh, vh) = 0) ⇒ (vh = 0) . 

wh∈Wh 

∀X ∈ R N , 〈AX, X〉N = a(Φ, Φ) ≥ α�Φ� 2 V ≥ c�X� 2 N , 

where 〈·, ·〉N denotes the Euclidean scalar product in RN , � · �N the associated norm and for X ∈ RN , 

X = (Xi)1≤i≤N , Φ = �N i=1 Xiϕi ∈ Vh. This result shows that A is positive definite. Conversely, if the 

matrix A is positive definite, for every X ∈ RN , with �X�N = 1, we have 〈AX, X〉N > 0. hence, we have 

∀Φ ∈ Vh , Φ �= 0 , 

� 

a(Φ, Φ) = 〈AX, X〉N = A X 

, 

�X�N 

X 

� 

�X�N 

�X� 2 N 

N 

≥ c�X� 2 N ≥ c�Φ� 2 V , 

noticing that the unit sphere of R N is compact and thus the continuous mapping X ↦→ 〈AX, X〉N is bounded 

by below by a constant c > 0. 

2. We have the following equivalence: 

we deduce 

X ∈ Ker(A) ⇔ ∀i , 

N� 

aijXj = 0 ⇔ ∀i , a(Φ, ϕi) = 0 

i=1 

⇔ sup a(Φ, v) = 0 , 

v∈Vh 

(c1h) ⇒ (∀Φ ∈ Vh , ( sup a(Φ, v) = 0) ⇒ (Φ = 0)) 

v∈Vh 

⇒ Ker(A) = {0} . 

Conversely, suppose Ker(A) = {0}. We consider a sequence (wn)n∈N ∈ Vh with �wn�W = 1 and such that 

a(wn, v) 

sup 

vh∈Vh �v�V 

≤ 1 

n . 

From the compacity of the unit sphere of Vh, we deduce that any subsequence also denoted (wn) tends 

towards w when n → ∞. The limit is such that �w�V = 1 and supv∈Vh a(w, v) = 0. This means that 

W ∈ Ker(A), where w = � N 

i=1 Wiϕi. Hence, W = 0 which is in contradiction with the assumption that 

�w�V = 1.


3. proceed similarly with A t . 

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 is 

a subspace of W , both of finite dimensions and we consider the discrete problem: 

⎧ 

⎪⎨ find uh ∈ Vh and ph ∈ Wh such that 

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


⎪⎩ 

b(uh, qh) = g(qh) , ∀qh ∈ Wh . 

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

problem. 

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

∃αh > 0 , inf 

sup 

uh∈Ker(Bh) vh∈Ker(Bh) 

a(uh, vh) 

�uh�V �vh�V 

� 

≥ αh , (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: 

and if 

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

∃βh > 0 , inf 

sup 

qh∈Wh vh∈Vh 

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

b(vh, qh) 

�vh�V �qh�W 

≥ βh . (4.25) 

�u − uh�V ≤ c1h inf �u − vh�V + c2h inf �p − qh�W , 

vh∈Vh 

qh∈Wh 


� 

with c1h = 1 + �a� 

� � 

1 + αh 

�b� 

� 

, c2h = βh 

�b� 

αh if Ker(Bh) �⊂ Ker(B) and c2h = 0 if Ker(Bh) ⊂ Ker(B) and 

with c3h = c1h �a� 

βh and c4h = 1 + �b� �a� 

+ c2h βh βh . 

�p − ph�W ≤ c3h inf �u − vh�V + c4h inf �p − qh�W , (4.27) 

vh∈Vh 

qh∈Wh 

Proof. It consists in using Theorem 4.8 while noticing that in finite dimension, the first condition on αh implies 

the first two conditions o the theorem (the injectivity and surjectivity are equivalent notions for an endomorphism 

in 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 βh�rh�X ≤ �b��u − vh�X .

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 that 

uh − wh ∈ Ker(Bh) and we have: 

a(uh − wh, yh) 

αh�uh − wh�x ≤ sup 

yh∈Ker(Bh) �yh�X 

a(uh − u, yh) + a(u − wh, yh) 

≤ sup 

yh∈Ker(Bh) 

�yh�X 

b(yh, p − ph) + a(u − wh, yh) 

≤ sup 

yh∈Ker(Bh) 

�yh�X 

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

The triangle inequality yields 

αh�uh − wh�X ≤ �a� �u − wh�X . 

�u − uh�X ≤ 

� 

1 + �a� 

� 

�u − wh�X . 

αh 

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

αh�uh − wh�X ≤ �a� �u − wh�X + �b� �p − qh�M . 

The triangle inequality yields 

� 

�u − uh�X ≤ 1 + �a� 

� 

�u − wh�X + 

αh 

�b� 

�p − qh�M . 

αh 

By noticing that the inequality �u − wh�X ≤ �u − vh�X + �rh�X can be written as follows: 

�u − wh�X ≤ 

� 

1 + �b� 

� 

�u − vh�X , 

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: 

βh�qh − ph�M ≤ �a� �u − uh�X + �b� �p − qh�M . 

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 basic 

boundary-value problems in one dimension. 

4.5.1 Dirichlet problem 

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

⎧ 

⎪⎨ − 

⎪⎩ 

d 

� 

ν 

dx 

du 

� 

+ µu = f , 

dx 

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

∀x ∈ Ω =]0, 1[ , 

, (4.28) 

βh


for ν and µ given functions of L ∞ (Ω) such that 

We pose 

ν(x) ≥ α > 0 , µ(x) ≥ 0 , a.e. in Ω . 

a(u, v) = 

� 1 

(νu 

0 

′ v ′ + µuv) dx , 

and the weak formulation of this problem consists in finding u ∈ V = H1 0 

∀v ∈ H 1 � 1 

0 (Ω) , a(u, v) = 

0 

(Ω) solving: 

fv dx . (4.29) 

The bilinear form a(·, ·) est continuous on V × V and V -elliptic, thus, for any f ∈ L 2 (Ω), the variational 

problem (4.29) is well-posed and has a unique solution u ∈ V . 

To construct an approximation uh of u, we consider the subspace Vh ⊂ V composed of piecewise affine 

continuous functions on intervals. Let introduce the step h = 1 

N+1 , for N ∈ N∗ given, and the points 

xn = nh, 0 ≤ n ≤ N + 1. The latter subdivide the interval ¯ Ω into N + 1 subintervals Kn = [xn, xn+1] of 

length 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 function 

(Ω) is (almost everywhere equal to) a continuous function on [0, 1] and satisfies the conditions 

v ∈ V = H1 0 

v(0) = v(1) = 0. Furthermore, any continuous function on Ω, piecewise C1 on intervals and such that 

v(0) = v(1) = 0 is a function of H1 0 (Ω). This leads to consider Vh as: 

Vh = {v ∈ C 0 ( ¯ Ω) , v(0) = v(1) = 0 , v|Kn ∈ P1 , 0 ≤ n ≤ N} . 

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

ϕi(xj) = δi,j , 1 ≤ j ≤ N , 

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

⎧ 

⎨ |x − xi| 

1 − 

ϕi = 

⎩ 

h 

0 otherwise 

∀x ∈ [xi−1, xi+1] , 

Hence, a function uh ∈ Vh can be uniquely determined using its coordinates un = uh(xn) in the basis 

(ϕn)1≤n≤N. The variational method consists in approaching the solution u ∈ V by the function uh = 

� N 

j=1 ujϕj ∈ Vh, where the (uj)1≤j≤N are solution of the linear system: 

N� 

a(ϕj, ϕi)uj = 

j=1 

� 1 

0 

fϕj dx , 1 ≤ i ≤ N . 

Actually, the linear system is relatively simple since the support of a function ϕj is the interval [xj−1, xj+1]. 

This leads to consider the system: 

⎧ 

⎪⎨ 

⎪⎩ 

a(ϕ1, ϕ1)u1 + a(ϕ2, ϕ1)u2 = 

. 

� 1 

0 

fϕ1 dx 

a(ϕi−1, ϕi)ui−1 + a(ϕi, ϕi)ui + a(ϕi+1, ϕi)ui+1 = 

. 

a(ϕN−1, ϕN)uN−1 + a(ϕN, ϕN)uN = 

� 1 

0 

fϕN dx 

� 1 

0 

fϕi dx

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: 

⎧ 

⎪⎨ 

⎪⎩ 

and ⎧⎪ ⎨ 

⎪⎨ 

µ − 3 

i+1/2 = 

h3 µ i+1/2 = 6 

h 3 

µ + 3 

i−1/2 = 

h3 ⎪⎩ 

νi+1/2 = 1 

h 

� xi+1 

ν(x) dx , 

xi 

� xi+1 

(xi+1 − x) 2 µ(x) dx , 

xi � xi+1 

xi � xi+1 

xi 

f − 2 

i+1/2 = 

h2 f + 2 

i+1/2 = 

h2 (xi+1 − x)(x − xi)µ(x) dx , 

(x − xi+1) 2 µ(x) dx , 

� xi+1 

xi � xi+1 

xi 

(xi+1 − x)f(x) dx , 

(x − xi+1)f(x) dx , 

yielding to the following linear system: 

⎧ � 

1 

h (ν1/2 + ν3/2) + h 

3 (µ+ 

� � 

1/2 + µ− 

3/2 

u1 + − 1 

h ν3/2 + h 

6 µ � 

3/2 

. 

� 

− 1 

h νi−1/2 + h 

6 µ � � 

1 

i−1/2 ui−1 + 

h (νi−1/2 + νi+1/2) + h 

3 (µ+ 

� 

− 1 

h νi+1/2 + h 

6 µ � 

i+1/2 

u2 = h + − 

(f 

2 1/2 + f3/2 ) , 

i−1/2 

+ µ− 

i+1/2 ) 

� 

ui+ 

ui+1 = h + − 

(f 

2 i−1/2 + fi+1/2 ) 

. 

� 

− 

⎪⎩ 

1 

h νN−1/2 + h 

6 µ � � 

1 

N−1/2 uN−1 + 

h (νN−1/2 + νN+1/2) + h 

3 (µ+ N−1/2 + µ− 

N+1/2 ) 

� 

uN 

= h + − 

(f 

2 N−1/2 + fN+1/2 ) 

The last issue that remains to be addressed concerns the right-hand side term. If the function f is an 

arbitrary function, the quantities f − 

+ 

i+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 

� � 

a + b 

g(x) dx ≈ (b − a)g . 

2 

a 

We proceed similarly for evaluating the quantities involving the functions ν and µ. It is important to 

carefully evaluate such quantities as it affects the stiffness matrix A of the system. 

The choice of the basis (ϕn)1≤n≤N as described above leads to a tridiagonal matrix A. Thus, efficient 

methods 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 

� 

ν 

dx 

du 

� 

+ µu = f , ∀x ∈ Ω =]0, 1[ , 

dx 

, (4.30) 

u ′ (0) = u ′ (1) = 0 .


for ν and µ given functions of L ∞ (Ω) such that 

ν(x) ≥ α > 0 , µ(x) ≥ 0 , a.e. in Ω . 

We have already seen that for any given f ∈ L 2 (Ω), this problem is well-posed and the solution u ∈ H 1 (Ω). 

Considering the subdivision of ¯ Ω into N + 1 subintervals Kn = [xn, xn+1] of length h, we introduce 

the space Vh ⊂ H 1 (Ω) defined as: 

Vh = {v ∈ C 0 ( ¯ Ω) , v|Kn ∈ P1 , 1 ≤ n ≤ N} . 

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

� x 

1 − if 1 ≤ x ≤ h 

ϕ0(x) = h 

0 otherwise 

⎧ 

⎨ |x − xi| 

1 − if xi−1 ≤ x ≤ xi+1 

ϕi(x) = h 

⎩ 

0 otherwise 

⎧ 

⎨ 1 − x 

1 − if 1 − h ≤ x ≤ 1 , 

ϕN+1(x) = h 

⎩ 

0 otherwise 

The variational approximation method consists in searching for the solution uh = �N+1 j=0 ujϕj ∈ Vh 

solving the linear system: 

⎧ 

⎪⎨ 

⎪⎩ 

� 1 

h (ν 1/2 + h 

3 (µ− 

1/2 

� 

� 

u0 + − 1 

h ν1/2 + h 

6 µ � 

1/2 

. 

� 

− 1 

h νi−1/2 + h 

6 µ � � 

1 

i−1/2 ui−1 + 

� 

− 1 

h νi+1/2 + h 

6 µ i+1/2 

u1 = h − 

(f 

2 1/2 ) , 

h (νi−1/2 + νi+1/2) + h 

3 (µ+ 

� 

i−1/2 

+ µ− 

i+1/2 ) 

� 

ui+ 

ui+1 = h + − 

(f 

2 i−1/2 + fi+1/2 ) 

. 

� 

− 1 

h νN+1/2 + h 

6 µ � � 

1 

N+1/2 uN + 

h νN+1/2 + h 

3 µ+ 

� 

N+1/2 

uN+1 = h + 

f 

2 N+1/2 

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

definite. 

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

u ′ h (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.

4.7. Exercises 75 

4.7 Exercises 

Exercise 4.1 Suppose Ω ⊂ Rd is an open set and f ∈ L2 (Ω) is a given function. We consider the 

following Neumann problem: ⎧ 

⎨ −∆u + u = f , on Ω 

⎩ 

∂u 

= 0 , 

∂n 

on ∂Ω . 


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

∀v ∈ H 1 � 

� 

(Ω) , ∇u · ∇v + uv dx = 

and that there is a unique solution to this problem in H 1 (Ω). 

Ω 

Ω 

fv dx , 

2. Sho wthat the solution u ∈ H 1 (Ω) of this problem verifies −∆u + u = f in D ′ (Ω). 

3. Suppose Ω is of class C 2 and thus u ∈ H 2 (Ω). In what sense u is a solution of the initial problem ?

76 Chapter 4. Weak formulation of elliptic problems

Chapter 4. Weak formulation of elliptic problems

You also want an ePaper? Increase the reach of your titles

Delete template?

Save as template?