1332 CHAPTER 39. WEAK SOLUTIONS

Here is why. It is obvious that A is in L (V,V ′) . It only remains to verify that it iscoercive.

A(u)(u) ≡∫

U

(α

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

dx

≥∫

Uδ |∇u(x)|2 + |u(x)|2 dx

≥ δ ||u||2H1(U)

This proves coercivity and verifies the claim.What has been obtained in the above example? This depends on how you choose V.

In Example 39.1.3 suppose U is a bounded open set with C0,1 boundary and V = H10 (U)

whereH1

0 (U)≡{

u ∈ H1 (U) : γu = 0}.

Also suppose f ∈ L2 (U) . Then you can consider F ∈V ′ by defining

F (v)≡∫

Uf (x)v(x)dx.

According to the Lax Milgram theorem and the verification of its conditions in Example39.1.3, there exists a unique solution to the problem of finding u ∈ H1

0 (U) such that for allv ∈ H1

0 (U) , ∫U

(α

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

dx =∫

Uf (x)v(x)dx (39.1.1)

In particular, this holds for all v ∈C∞c (U) . Thus for all such v,∫

U

(−(α

i j (x)u,i (x)), j +u(x)− f (x)

)v(x)dx = 0.

Therefore, in terms of weak derivatives,

−(α

i ju,i), j +u = f

and since u ∈ H10 (U) , it must be the case that γu = 0 on ∂U. This is why the solution to

39.1.1 is referred to as a weak solution to the boundary value problem

−(α

i j (x)u,i (x)), j +u(x) = f (x) , u = 0 on ∂U.

Of course you then begin to ask the important question whether u really has two derivatives.It is not immediately clear that just because −

(α i j (x)u,i (x)

), j ∈ L2 (U) it follows that the

second derivatives of u exist. Actually this will often be true and is discussed somewhat inthe next section.

Next suppose you choose V = H1 (U) and let g ∈ H1/2 (∂U). Define F ∈V ′ by

F (v)≡∫

Uf (x)v(x)dx+

∫∂U

g(x)γv(x)dµ.