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Numerical Analysis of a Finite Element/Volume Penalty Method 175
Example 2. The simplest example of penalty formulation one may think about
is the following: the constraint to vanish on a subdomain O⊂⊂Ω is handled
by minimizing the functional
J ε (v) = 1 ∫ ∫
|∇v| 2 − fv + 1 ∫
u 2 . (7)
2 Ω
Ω 2ε O
Example 1 (which is a one-dimensional version of this situation) suggests that
|u ε − u| behaves like ε 1/4 . If we admit this convergence rate, Proposition 6
provides an estimate in h 1/2 + ε 1/8 . This estimate is optimal in h: the natural
space discretization order is obtained if ε is small enough (ε = h 4 in the
present case).
Symmetrically, the natural order in ε can be recovered if h is small enough.
Indeed, if we admit that u ε can be approximated at the same order as u over
Ω, which is 1/2, then the choice ε = h 4/3 in Proposition 7 gives
|u ε h − u| ≤ C
ε 1/2 ε3/4 + ε 1/4 = O(ε 1/4 ).
Notice that if we replace u 2 by u 2 + |∇u| 2 in the integral over O in (7),
the assumptions of Corollary 1 are fulfilled, so that convergence holds at the
first order in ε. As a consequence, |u − u ε h | is bounded by C(h1/2 + ε 1/2 )(by
Proposition 6), which suggests the choice ε = h.
3 Model Problem
This section is dedicated to a special situation, which can be seen as a scalar
version of the rigidity constraint in a Stokes flow.
We introduce a domain Ω ⊂ R 2 (smooth, or polygonal and convex), and
O⊂⊂Ω which we suppose circular (see Remark 4, at the end of this paper,
for extensions to more general situations). We consider the following problem:
⎧
−△u = f in Ω \ O,
⎪⎨ u =0 on∂Ω,
u = U on ∂O,
(8)
∫
∂u ⎪⎩
∂n =0,
∂O
where U is an unknown constant, and f ∈ L 2 (Ω \ O). The scalar field u can
be seen as a temperature, and O as a zone with infinite conductivity.
Definition 1. We say that u is a weak solution to (8) if u ∈ V = H0 1 (Ω),
there exists U ∈ R such that u = U almost everywhere in O, and
∫
∫
∇u ·∇v = fv ∀v ∈ D(Ω).
Ω
Ω