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Modélisation, analyse mathématique et simulations numériques de ...

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tel-00656013, version 1 - 3 Jan 2012<br />

150 Asymptotic analysis of blood flow in stented arteries<br />

On the other hand, for any test function φ ∈ H 1 0 (Ωε):<br />

<br />

Ωε<br />

Lk ˜ R ε 0φdx = Ĉk<br />

<br />

Ωε\Ω0<br />

φdx+Mk<br />

<br />

Γ0<br />

φdx.<br />

Then, using Cauchy-Schwarz and Poincaré like inequalities, we obtain the upper bound:<br />

<br />

<br />

<br />

<br />

≤ c | Ĉk|ε+|Mk| √ <br />

ε ˜ R ε 0H1 (Ωε), (5.3.9)<br />

ik ˜ R ε 0 2 L 2 (Ωε) +∇˜ R ε 0 2 L 2 (Ωε)<br />

where c is a non negative constant <strong>de</strong>pending on the Poincaré inequality. And |Mk| is<br />

controlled as follows:<br />

|Mk| =<br />

<br />

<br />

Ĉkr (2−e r −e−r <br />

<br />

) <br />

|ik(er −e−r )|<br />

≤ |Ĉk|<br />

√k<br />

≤ 2| Ĉk|.<br />

<br />

2<br />

|er −e−r | +1<br />

<br />

(5.3.10)<br />

Finally, combining (5.3.8)-(5.3.10), we g<strong>et</strong> the H 1 -error estimate.<br />

For the L 2 error, we use the concept of a very weak solution. Namely, one solves the<br />

dual problem: for a given φ ∈ L 2 (Ω0), φ being x1 periodic on Γin ∪Γout, find ˆv ∈ H 2 (Ω0)<br />

such that ⎧ ⎨<br />

⎩<br />

Lk ˆv = φ in Ω0,<br />

ˆv = 0 on Γ1 ∪Γ0,<br />

ˆv is x1-periodic on Γin ∪Γout.<br />

Then, consi<strong>de</strong>ring the L 2 (Ω0) scalar product . , . , and using the Green formula:<br />

(R ε 0<br />

<br />

, φ) =<br />

Ω0<br />

R ε 0 Lkˆv = −ik<br />

<br />

Ω0<br />

R ε 0<br />

=<br />

<br />

ˆv +<br />

<br />

Ω0<br />

ˆv , ∂Rε 0<br />

∂n<br />

∇R ε <br />

0∇ˆv −<br />

<br />

Γin∪Γout<br />

−<br />

∂Ω0<br />

R ε 0<br />

∂ˆv<br />

∂n<br />

<br />

R ε 0 , ∂ˆv<br />

∂n<br />

<br />

Γ0∪Γ1<br />

(5.3.11)<br />

, (5.3.12)<br />

where the brack<strong>et</strong>s refer to the dual product in H−1 , H1 (∂Ω0) and the rest of the<br />

products are in L2 either on Γ0 or in Ω0. Then one computes:<br />

|(R ε 0 , φ)| ≤ R ε <br />

<br />

0L2 <br />

∂ˆv <br />

<br />

(Γ0) ∂n<br />

L 2 (Γ0)<br />

≤ √ ε∇R ε 0 L 2 (Ωε\Ω0) φ L 2 (Ω0)<br />

≤ √ ε∇R ε 0 L 2 (Ωε) φ L 2 (Ω0)<br />

≤ ε 3<br />

2 φ L 2 (Ω0) .

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