Chapter 39

Weak Solutions39.1 The Lax Milgram Theorem

The Lax Milgram theorem is a fundamental result which is useful for obtaining weak so-lutions to many types of partial differential equations. It is really a general theorem infunctional analysis.

Definition 39.1.1 Let A ∈ L (V,V ′) where V is a Hilbert space. Then A is said to becoercive if

A(v)(v)≥ δ ||v||2

for some δ > 0.

Theorem 39.1.2 (Lax Milgram) Let A ∈L (V,V ′) be coercive. Then A maps one to oneand onto.

Proof: The proof that A is onto involves showing A(V ) is both dense and closed.Consider first the claim that A(V ) is closed. Let Axn→ y∗ ∈V ′. Then

δ ||xn− xm||2V ≤ ||Axn−Axm||V ′ ||xn− xm||V .

Therefore, {xn} is a Cauchy sequence in V. It follows xn→ x∈V and since A is continuous,Axn→ Ax. This shows A(V ) is closed.

Now let R : V → V ′ denote the Riesz map defined by Rx(y) = (y,x) . Recall that theRiesz map is one to one, onto, and preserves norms. Therefore, R−1 (A(V )) is a closed sub-space of V. If there R−1 (A(V )) ̸= V, then

(R−1 (A(V ))

)⊥ ̸= {0} . Let x ∈(R−1 (A(V ))

)⊥and x ̸= 0. Then in particular,

0 =(x,R−1Ax

)= R

(R−1 (A(x))

)(x) = A(x)(x)≥ δ ||x||2V ,

a contradiction to x ̸= 0. Therefore, R−1 (A(V )) =V and so A(V ) = R(V ) =V ′.Since A(V ) is both closed and dense, A(V ) =V ′. This shows A is onto.If Ax = Ay, then 0 = A(x− y)(x− y)≥ δ ||x− y||2V , and this shows A is one to one. This

proves the theorem.Here is a simple example which illustrates the use of the above theorem. In the example

the repeated index summation convention is being used. That is, you sum over the repeatedindices.

Example 39.1.3 Let U be an open subset of Rn and let V be a closed subspace of H1 (U) .Let α i j ∈ L∞ (U) for i, j = 1,2, · · · ,n. Now define A : V →V ′ by

A(u)(v)≡∫

U

(α

i j (x)u,i (x)v, j (x)+u(x)v(x))

dx.

Suppose also thatα

i jviv j ≥ δ |v|2

whenever v ∈ Rn. Then A maps V to V ′ one to one and onto.

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