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54 2. Finite Element Method for Poisson Equation

The weak derivative is well-defined because it is unique: Let v 1 ,v 2 ∈
2
L (Ω) be two weak derivatives of u. It follows that

∞
(v 1 − v 2 ) ϕdx =0 for all ϕ ∈ C (Ω) .
0
Ω
2
Since C (Ω) is dense in L (Ω), we can furthermore conclude that
∞
0

2
(v 1 − v 2 ) ϕdx =0 for all ϕ ∈ L (Ω) .
Ω
If we now choose specifically ϕ = v 1 − v 2 ,we obtain

2
v 1 − v 2 = (v 1 − v 2 )(v 1 − v 2 ) dx =0 ,
0
Ω
k ¯
and v 1 = v 2 (a.e.) follows immediately. In particular, u ∈ C (Ω) has weak
α
derivatives ∂ u for α with |α|≤ k, and the weak derivatives are identical
to the classical (pointwise) derivatives.
Also the differential operators of vector calculus can be given a weak
definition analogous to Definition 2.9. For example, for a vector field q
2
2
with components in L (Ω),v ∈ L (Ω) is the weak divergence v = ∇· q if
for all ϕ ∈ C (Ω)
∞
0

vϕ dx = − q ·∇ϕdx.
Ω Ω
1
The correct choice of the space V is the space H (Ω), which will be
0
defined below. First we define

2
1
H (Ω) := u :Ω → R u ∈ L (Ω) ,u has weak derivatives

(2.17)
2
∂ i u ∈ L (Ω) for all i =1,... ,d .
1
A scalar product on H (Ω) is defined by

u, v := u(x)v(x) dx + ∇u(x) ·∇v(x) dx (2.18)
1
Ω Ω
with the norm
1/2
(
2
2
u 1 := u, u = |u(x)| dx + |∇u(x)| dx (2.19)
1
Ω Ω
induced by this scalar product.
The above “temporary” definition (2.7) of V takes care of the boundary
condition u =0 on ∂Ω by conditions for the functions. I.e. we want to
choose the basic space V analogously as:

1
1
H (Ω) := u ∈ H (Ω) u =0 on ∂Ω . (2.20)

0
1
1
Here H (Ω) and H (Ω) are special cases of so-called Sobolev spaces.
0
1
d
For Ω ⊂ R , d ≥ 2, H (Ω) may contain unbounded functions. In par-
ticular, we have to examine carefully the meaning of u| ∂Ω (∂Ωhas the