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Numerical Analysis of a Finite Element/Volume Penalty Method 171
Proof. This is a direct consequence of the identity which we obtain by subtracting
(4) and (2):
〈ξ,v〉− 1 ε b(uε ,v)=a(u − u ε ,v)
∀v ∈ V
which yields ‖ξ − ξ ε ‖ V ′ ≤ C |u − u ε |.
Let us now establish the first order convergence, provided an extra compatibility
condition between b(·, ·) andξ is met.
Proposition 3. Under the assumptions (1), we assume, in addition, that
there exists ˜ξ ∈ V such that
Then |u ε − u| = O(ε).
〈ξ,v〉 = b(˜ξ,v) ∀v ∈ V.
Proof. First of all, notice that it is possible to pick ˜ξ in K ⊥ (if not, we project
it onto K ⊥ ). Now following the idea which is proposed in [Bab73] in a slightly
different context (see the proof of Theorem 3.2 therein), we introduce
and we develop
R ε (v) = 1 2 a(u − v, u − v)+ 1 2ε b(ε˜ξ − v, ε˜ξ − v)
R ε (v) = 1 2 a(u, u)+ ε 2 b(˜ξ, ˜ξ)+ 1 2 a(v, v)+ 1 b(v, v) − a(u, v) − b(˜ξ,v).
2ε
As b(˜ξ,v) =〈ξ,v〉 and −a(u, v)−〈ξ,v〉 = −〈ϕ, v〉, the functional R ε is equal to
J ε up to a constant. Therefore, minimizing R ε or minimizing J ε are equivalent
tasks. Let us now introduce w = ε˜ξ + u. It comes
R ε (w) = ε2
2 a(˜ξ, ˜ξ) + 0 because u ∈ K =kerb,
so that |R ε (w)| ≤Cε 2 .Asu ε minimizes R ε ,
0 ≤ R ε (u ε )= 1 2 a(u − uε ,u− u ε )+ 1 2ε b(ε˜ξ − u ε ,ε˜ξ − u ε ) ≤ Cε 2 ,
from which we deduce, as a(·, ·) is elliptic, |u − u ε | = O(ε).
Corollary 1. Under the assumptions (1), we assume, in addition, that b(·, ·)
can be written b(u, v) =(Bu,Bv), where B is a linear continuous operator
onto a Hilbert space Λ, with closed range. Then |u ε − u| = O(ε).