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48 2. Finite Element Method for Poisson Equation

∞
The mapping a defines a scalar product on C (Ω) that induces the norm
0
1/2
" 2
u a := a(u, u)= |∇u| dx (2.6)
Ω
(see Appendix A.4 for these notions). Most of the properties of a
scalar product are obvious. Only the definiteness (A4.7) requires further
considerations. Namely, we have to show that

a(u, u)= (∇u ·∇u)(x) dx =0 ⇐⇒ u ≡ 0 .
Ω
To prove this assertion, first we show that a(u, u) = 0 implies ∇u(x)= 0
for all x ∈ Ω. To do this, we suppose that there exists some point ¯x ∈ Ω
2
such that ∇u(¯x) =0. Then (∇u ·∇u)(¯x)= |∇u| (¯x) > 0. Because of
the continuity of ∇u, a small neighbourhood G of ¯x exists with a positive
2
measure |G| and |∇u|(x) ≥ α> 0 for all x ∈ G.Since |∇u| (x) ≥ 0 for all
x ∈ Ω, it follows that

2 2
|∇u| (x) dx ≥ α |G| > 0 ,
Ω
which is in contradiction to a(u, u)= 0. Consequently, ∇u(x)= 0holds
for all x ∈ Ω; i.e., u is constant in Ω. Since u(x) = 0 for all x ∈ ∂Ω, the
assertion follows.
∞
Unfortunately, the space C (Ω) is too small to play the part of the basic
0
space because the solution u does not belong to C (Ω) in general. The
∞
0
identity (2.4) is to be satisfied for a larger class of functions, which include,
as an example for v,the solution u and the finite element approximation
to u to be defined later.
For the present we define as the basic space V ,

¯
V := u :Ω → R u ∈ C(Ω) ,∂ i u exists and is piecewise

(2.7)
continuous for all i =1,... ,d, u =0 on ∂Ω .
To say that ∂ i u is piecewise continuous means that the domain Ω can be
decomposed as follows:
#
¯
¯
Ω= Ω j ,
j
with a finite number of open sets Ω j ,withΩ j ∩ Ω k = ∅ for j = k, and ∂ i u
¯
is continuous on Ω j and it can continuously be extended on Ω j .
Then the following properties hold:
• a is a scalar product also on V ,
∞
• C (Ω) ⊂ V ,
0
∞
• C (Ω) is dense in V with respect to · a; i.e., for any u ∈ V (2.8)
0
a sequence (u n ) n∈N in C (Ω) exists such that u n −u a → 0
∞
0
for n →∞,