The Lax-Milgram Theorem

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Definition 1.

A bilinear form a(·, ·) on a normed linear space H is said to be bounded

or continuous, if ∃C < ∞ such that

|a(v, w)| ≤ C‖v‖H‖w‖H ∀v, w ∈ H , and coercive on V ⊂ H if

∃α > 0 such that a(v, v) ≥ α‖v‖2H ∀v ∈ V .

Propositon 1 Let H be a Hilbert space, and suppose that a(·, ·) is a

symmetric bilinear form that is continuous on H and coercive on V ⊂ H .

Then, V, a(·, ·) is a Hilbert space.

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Proof

Because a(·, ·) is coercive → a(·, ·) is an inner product on V .

Next, let ‖v‖E =√

a(v, v) and suppose that vn is a Cauchy sequence

in (V, ‖ · ‖E). By coercivity, vn is also Cauchy sequence in (H, ‖ · ‖H).

Since H is complete, ∃v ∈ H : vn → v in the ‖ · ‖H norm. Since V is

closed in H, v ∈ V .

Since a(·, ·) is bounded → ‖v − vn‖H ≤ √c1‖v − vn‖H . Hense, vn → v

in the ‖ · ‖E norm and (V, ‖ · ‖E) is complete.

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A symmetric variational problem

If the following conditions are valid

1. H, (·, ·) is a Hilbert space,

2. V is closed subspace of H,

3. a(·, ·) is bounded, symmetric form that is coercive on V ,

then the symmetric variational problem is the following:

Given F ∈ V ′ find u ∈ V such that a(u, v) = F (v) ∀v ∈ V.

Theorem

Suppose that conditions (1)-(3) above hold. Then ∃!u ∈ V solvinga(u, v) = F (v) ∀v ∈ V.

Proof

Proposition 1 implies, that a(·, ·) is an inner product on V and V, a(·, ·) isa Hilbert space. Then we apply the Riesz Theorem. 2

4

The Ritz-Galerkin approximation

Given a finite-dimensional subspace Vh ⊂ V and F ∈ V ′, find uh ∈ Vh

such that a(uh, v) = F (v) ∀v ∈ Vh.

Theorem

Under the conditions (1)-(3) there exists a unique uh that solves

a(uh, v) = F (v) ∀v ∈ Vh.

Proof

Vh, a(·, ·) is a Hilbert space, and F ∈ V ′h. Then proof follows from Riesz

theorem.

Fundamental Orthogonality

Let u, uh be the solutions to a(u, v) = F (v) and a(uh, v) = F (v),

respectively, then a(u − uh, v) = 0 ∀v ∈ Vh.

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The Ritz method

In the symmetric case, uh minimizes the quadratic functional

Q(v) = a(v, v) − 2F (v) ∀v ∈ Vh.

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Formulation of nonsymmetric variational problems

1. H, (·, ·) is a Hilbert space.

2. V is closed subspace of H .

3. a(·, ·) is a bilinear form on V , not necessarily symmetric.

4. a(·, ·) is continuous (bounded) on V .

5. a(·, ·) is coercive on V .

then the nonsymmetric variational problem is the following:

Given F ∈ V ′ find u ∈ V such that a(u, v) = F (v) ∀v ∈ V.

Galerkin approximation problem

Given a finite-dimensional subspace Vh ⊂ V and F ∈ V ′, find uh ∈ Vh

such that a(uh, v) = F (v) ∀v ∈ Vh.

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The Lax-Milgram Theorem

We would like to prove the existence and uniqueness of the solution of the(nonsymmetric) variational problem.

Lemma. Contraction mapping Principle

Given a Banach space V and a mapping T : V → V , satisfying

||Tv1 − Tv2|| ≤ M ||v1 − v2|| ∀v1, v2 ∈ V and fixed M, 0 ≤ M < 1,there ∃!u ∈ V such that

u = Tu,

or the contraction mapping T has a unique fixed point u.

Proof

1) We show uniqueness. Let Tv1 = v1 and Tv2 = v2. Since T is acontraction mapping, ||Tv1 − Tv2|| ≤ M ||v1 − v2|| for someM, 0 ≤ M < 1. But ||Tv1 − Tv2|| = ||v1 − v2||, therefore,

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||v1 − v2|| ≤ M ||v1 − v2||, and this implies ||v1 − v2|| = 0 (otherwise

1 ≤ M ). Then v1 = v2.

2) Existence. We take v0 ∈ V and define

v1 = Tv0, v2 = Tv − 1, ..., vk+1 = Tvk. Note that

||vk+1 − vk|| = ||Tvk − Tvk−1|| ≤ M ||vk − vk−1||. By induction,

||vk − vk−1|| ≤ Mk−1||v1 − v0||. Therefore, ∀N > n

||vN − vn|| = ||N

∑

k=n+1

vk − vk−1||

≤ ||v1 − v0||N

∑

k=n+1

Mk−1

≤ Mn

1 − M||v1 − v0||

=Mn

1 − M||Tv0 − v0||,

(1)

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from what follows that vn is a Cauchy sequence. Since V is complete

and limn→∞ vn =: v, we have

v = limn→∞

vn+1

= limn→∞

Tvn

= T ( limn→∞

vn) (T is continuous)

= Tv,

(2)

or there exists a fixed point.

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The Lax-Milgram Theorem

Given a Hilbert space V, (·, ·), a continuous, coercive bilinear form a(·, ·)and a continuous linear functional F ∈ V ′, there exists a unique u ∈ V

such that

a(u, v) = F (v) ∀v ∈ V. (3)

Proof

For any u ∈ V we define a functional Au(v) = a(u, v) ∀v ∈ V . Au is

linear since

Au(αv1 + βv2) = a(u, αv1 + βv2))

= αa(u, v1) + βa(u, v2)

= αAu(v1) + βAu(v2) ∀v1, v2 ∈ V, α, β ∈ R.

Au is continuous, because ∀v ∈ V

|Au(v)| = |a(u, v)| ≤ C||u||||v|| ∀v ∈ V.

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Therefore,

||Au(v)||V ′ = supv 6=0

Au(v)

||v|| ≤ C||u|| < ∞.

Thus, Au ∈ V ′. Similarly one can show that the mapping u → Au is a

linear map V → V ′.

By Riesz representation theorem, for ∀ϕ ∈ V ′ there ∃!τϕ ∈ V such that

ϕ(v) = (τϕ, v) ∀v ∈ V . We must find a unique u such that

Au(v) = F (v) ∀v ∈ V , in other words, we want to find a unique u

such that Au = f in V ′ or τAu = τF in V , since τ : V ′ → V is

one-to-one mapping.

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Now we use Contraction Mapping Principle. We want to find ρ 6= 0 such

that the mapping T : V → V is a contraction mapping, where T is

defined by:

Tv := v − ρ(τAv − τF ) ∀v ∈ V (4)

If T is a contraction mapping, then by Contraction Mapping Principle

there ∃!u ∈ V such that Tu = u − ρ(τAu − τF ) = u, that is

ρ(τAu − τF ) = 0 or τAu = τF . It remains to show that there ∃ρ 6= 0.

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For ∀v1, v2 ∈ V let v = v1 − v2, then

||Tv1 − Tv2||2 = ||v1 − v2 − ρ(τAv1 − τAv2)||2

= ||v − ρ(τAv)||2

= ||v||2 − 2ρ(τAv, v) + ρ2||τAv||2

= ||v||2 − 2ρAv(v) + ρ2Av(τAv)

(definition of τ , (τAv, v) = Av(v))

= ||v||2 − 2ρa(v, v) + ρ2a(v, τAv)

(definition of A)

≤ ||v||2 − 2ρα||v||2 + ρ2C||v|| · ||τAv||(coercivity and continuity of A)

≤ (1 − 2ρα + ρ2C2)||v||2

(A bounded, τ isometric and ||τAv|| = ||Av|| ≤ C||v||)≤ (1 − 2ρα + ρ2C2)||v1 − v2||2

= M2||v1 − v2||2.14

We need 1 − 2ρα + ρ2C2 < 1 for some ρ. If we choose ρ ∈ (0, 2α/C2)

then M < 1 and the proof is complete.

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Corollary

The nonsymmetric variational problem has a unique solution.

Theorem (Cea)

Suppose that conditions for nonsymmetric variational problem hold and

that u solves a(u, v) = F (v) ∀v ∈ V. For the finite element variational

problem a(uh, v) = F (v) ∀v ∈ Vh we have

||u − uh||V ≤ C

αminv∈Vh

||u − v||V ,

where C is the continuity constant and α is the coercivity constant of

a(·, ·).

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Proof

Since a(u, v) = F (v) ∀v ∈ V and a(uh, v) = F (v) ∀v ∈ Vh we

have a(u − uh, v) = 0 ∀v ∈ Vh. For ∀v ∈ Vh

α||u − uh||2V ≤ a(u − uh, u − uh)

= a(u − uh, u − v) + a(u − uh, v − uh)

= a(u − uh, u − v)(since v − uh ∈ Vh and use orthogonality)

≤ C||u − uh||V · ||u − v||V .

Hence, ||u − uh||V ≤ Cα ||u − v||V . Therefore,

||u − uh||V ≤ C

αinf

v∈Vh

||u − v||V

=C

αminv∈Vh

||u − v||V (since Vh is closed).

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16

Applications of the Lax-Milgram theorem

A problem with Neumann boundary conditions We consider Poisson’sequation in 1D with an absorption term together with Neumann boundaryconditions:

−u′′ + u = f in Ω

∂xu|x=0,1 = 0,(5)

where Ω ⊂ R. The variational form is:

a(v, w) =

∫

Ω

(v′ · w′ + vw)dx,

L(v) =

∫

Ω

fvdx,

V = v :

∫

Ω

(|v′|2 + v2)dx < ∞.

(6)

We will verify that assumptions of the Lax-Milgram theorem are satisfied

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with these choices.

1. V has natural scalar product and norm defined as

(v, w)V :=

∫

Ω

(v′w′ + vw)dx,

||v||V :=

∫

Ω

(|v′|2 + v2)1/2dx.

(7)

Then, we see that V is complete, and therefore, V is Hilbert space.

2. a(v, v)V is V -elliptic-coercive and continuous - trivially holds.

3. L(·) is continuous. We note, that

|L(v)| ≤ ||f ||L2(Ω)||u||L2(Ω) ≤ ||f ||L2(Ω)||u||V , (8)

which means that we can take constant in definition of continuity as

k = ||f ||L2(Ω).

Then we can apply Lax-Milgram theorem to (5).

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The spaces H1(Ω) and H10 (Ω)

The space defined in (13) has a special notation

H1(Ω) = v :

∫

Ω

(|v′|2 + v2)dx < ∞, (9)

while the scalar product and norm are defined by

(v, w)H1(Ω) =

∫

Ω

(v′w′ + vw)dx,

||v||H1(Ω) :=

∫

Ω

(|v′|2 + v2)1/2dx

(10)

We also use subspase H10 (Ω) of H1(Ω):

H10 (Ω) = v ∈ H1(Ω) : v = 0 on Γ. (11)

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Poisson’s equation in 1D with Dirichlet boundary conditions

We consider Poisson’s equation in 1D with with Dirichlet boundaryconditions:

−u′′ = f in Ω

u(0) = u(1) = 0,(12)

where Ω ⊂ R. The variational form is:

a(v, w) =

∫

Ω

v′ · w′dx,

L(v) =

∫

Ω

fvdx,

V = H10 (Ω).

(13)

In this case the V -ellipticity of a does not follow automatically from thedefinition of norm V = H1

0 (Ω) as above (a(v, w) does not contain theterm

∫

Ωv2dx). To verify the ellipticity, we use the

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Poincare-Friedrichs inequality

||v||L2(0,1) ≤ 2(v(0)2 + ||v′||2L2(0,1)) (14)

Proof

v2(x) = (v(0) +

∫ x

0

v′(y)dy)2 ≤ 2(v2(0) +

∫ 1

0

(v′(y))2dy) (15)

for 0 ≤ x ≤ 1 ( we use integrating, Caushy’s inequality and the fact that(a + b)2 ≤ 2(a2 + b2)). 2

For functions v ∈ H1(Ω) with v(0) = v(1) = 0, or ||v||L2(Γ) = 0,

Poincare-Friedrichs inequality means

||v||H1(Ω) = ||v′||2L2(0,1)+||v||2L2(0,1) ≤ (1+C)||v′||2L2(0,1) = (1+C)a(v, v),

(16)which proves V -ellipticity with constant in definition k1 = 1

1+C . In 1D

constant C = 2.

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Continuity of a(·, ·) and L(·) follow exactly as for the case of Neumann

boundary conditions.

Remains to show that H10 (Ω) is a well defined Hilbert space, or that a

function in H10 (Ω) has well defined values on the boundary. In general it

is impossible to uniquely define the boundary values of v ∈ L2(Ω),

because v changes only close to the boundary. However, if we change

L2(Ω) to H1(Ω), then the trace of v ∈ H1(Ω) on the boundary is well

defined. This expressed in the trace inequality:

Theorem

If Ω is a bounded domain with boundray Γ, then there is a constant C

such that ∀v ∈ H1(Ω)

||v||L2(Γ) ≤ C||v||H1(Ω) (17)

Proof

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We take ϕ = |x|2

2 such that ϕ′′ = 1. The we use the fact that

∫ 1

0

v2ϕ′′dx = v2∂xϕ|10 −∫ 1

0

2vv′ϕ′dx (18)

and choose ∂xϕ = 1 to get from (18):∫ 1

0

v2dx = v2|10 −∫ 1

0

2vv′ϕ′dx, (19)

therefore,

v2|10 ≤∫ 1

0

2vv′ϕ′dx +

∫ 1

0

v2dx, (20)

and applying Poincare-Friedrichs inequality we get

||v2||L2(0,1) ≤ C(

∫ 1

0

v′2dx +

∫ 1

0

v2dx) = C||v||H1(Ω). (21)

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