Modélisation, analyse mathématique et simulations numériques de ...

tel-00656013, version 1 - 3 Jan 2012
5.3 Time Fourier analysis and boundary layer approximations 149
in the second case one should lift the Dirichlet error [40]. Here, we chose to extend û0,k by
a linear function in Ωε \Ω0:
û0,k in Ω0,
û0,k :=
, where Mk :=
Mkx2 in Ωε \Ω0,
∂
û0,k(x1,0) =
∂x2
Ĉkr(2−e r −e−r )
ik(e−r −er .
)
5.3.2 Zero order error estimates
We detail here the error estimates. Identical proofs should also be used for higher order
approximations below: we detail here every step. Denote χΩ the characteristic function of
the domain Ω, δΓ0 the Dirac measure concentrated on Γ0 .
Proposition 1. There exist two positive constants c1 and c2, depending only of the mode
Ĉk and the Sobolev’s inequalities, such that:
√
ε , ûǫ,k −û0,kL 2 (Ω0) ≤ c2ε. (5.3.4)
ûǫ,k −û0,k H 1 (Ωε) ≤ c1
Proof. The first part of the proof is based on standard a priori estimates. The existence
and uniqueness of ûǫ,k are well known and derive from the Lax-Milgram theorem. We focus
on the error, namely we set R ε 0 := ûǫ,k−û0,k. Since the extension û0,k of û0,k in the rough
domain satisfies: ⎧⎨
⎩
Lk û0,k = ĈkχΩ0 +ikMkx2χ Ωε\Ω0 in Ωε,
û0,k = 0 on Γ1,
û0,k = Mkx2
on Γε.
Then the zeroth order error solves:
⎧
⎨ Lk R
⎩
ε 0 = ĈkχΩε\Ω0 −ikMkx2χ Ωε\Ω0 in Ωε,
Rε 0 = 0 on Γ1,
= −Mkx2
on Γε.
R ε 0
(5.3.5)
(5.3.6)
We remark that a part of the error comes from the source term localized in Ωε \Ω0, and
another part comes from the non homogeneous boundary term on Γε. We set the lift:
Then: ⎧ ⎨
⎩
s = −Mkx2χ Ωε\Ω0 , ˜ R ε 0 = R ε 0 −s.
Lk ˜ Rε 0 = ĈkχΩε\Ω0 +MkδΓ0 in Ωε,
˜R ε 0 = 0 on Γ1,
˜R ε 0 = 0 on Γε,
(5.3.7)
where the derivatives are computed in the sense of distributions. Then, on the one hand,
using Poincaré inequality, we have:
Lk ˜ R ε 0 ˜ Rε 0dx
2
= ik ˜ R ε 0 2 L2 (Ωε) +∇˜ R ε 0 2 L2
(Ωε) 2
Ωε
= k 2 ˜ R ε 0 4 L 2 (Ωε) +∇˜ R ε 0 4 L 2 (Ωε)
≥ c ˜ R ε 0 4 H 1 (Ωε) .
(5.3.8)