39.1. THE LAX MILGRAM THEOREM 1333

Everything works the same way and you get the existence of a unique u ∈H1 (U) such thatfor all v ∈ H1 (U) ,∫

U

(α

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

dx =∫

Uf (x)v(x)dx+

∫∂U

g(x)γv(x)dµ (39.1.2)

is satisfied. If you pretend u has all second order derivatives in L2 (U) and apply the diver-gence theorem, you find that you have obtained a weak solution to

−(α

i ju,i), j +u = f , α

i ju,in j = g on ∂U

where n j is the jth component of n, the unit outer normal. Therefore, u is a weak solutionto the above boundary value problem.

The conclusion is that the Lax Milgram theorem gives a way to obtain existence anduniqueness of weak solutions to various boundary value problems. The following theoremis often very useful in establishing coercivity. To prove this theorem, here is a definition.

Definition 39.1.4 Let U be an open set and δ > 0. Then

Uδ ≡{

x ∈U : dist(x,UC)> δ

}.

Theorem 39.1.5 Let U be a connected bounded open set having C0,1 boundary such thatfor some sequence, ηk ↓ 0,

U = ∪∞k=1Uηk (39.1.3)

and Uηk is a connected open set. Suppose Γ⊆ ∂U has positive surface measure and that

V ≡{

u ∈ H1 (U) : γu = 0 a.e. on Γ}.

Then the norm |||·||| given by

|||u||| ≡(∫

U|∇u|2 dx

)1/2

is equivalent to the usual norm on V.

Proof: First it is necessary to verify this is actually a norm. It clearly satisfies all theusual axioms of a norm except for the condition that |||u|||= 0 if and only if u = 0. Supposethen that |||u|||= 0. Let δ 0 = ηk for one of those ηk mentioned above and define

uδ (x)≡∫

B(0,δ )u(x−y)φ δ (y)dy

where φ δ is a mollifier having support in B(0,δ ) . Then changing the variables, it followsthat for x ∈Uδ 0

uδ (x) =∫

B(x,δ )u(t)φ δ (x− t)dt =

∫U

u(t)φ δ (x− t)dt