Modélisation, analyse mathématique et simulations numériques de ...

tel-00656013, version 1 - 3 Jan 2012
108 An Asymptotic Preserving scheme
Hence, we have
δ n+1
j = δ n+1/2
j
1 +
1−
n+1/2
∂vR(uj ,η)
β
Therefore under the assumption (4.2.2), we set for all s ≥ 0
g(s) = 1+ 1− β0
s e
β
−s ,
β∆t
e
ε
−β∆t/ε .
for which we easily show that for all s ∈ R + , we have that e −s ≤ g(s) ≤ e −β0s/β . Hence,
for s = ∆t/ε δ n+1 L 1 ≤ e −β0∆t/ε δ n+1/2 L 1 . (4.4.4)
Finally, gathering (4.4.3) and (4.4.4), we obtain that there exists a constant C1 > 0 depending
only on a, TV(u 0 ) and TV(v 0 ) such that
δ n+1 L 1 ≤ e −β0∆t/ε [δ n L 1 + C1∆t] .
By induction, we easily get
δ n L 1 ≤ e −β0t n /ε δ 0 L 1 + Ca∆t
e−β0∆t/ε . (4.4.5)
1−e −β0∆t/ε
To conclude we only observe that xe −x ≤ 1−e −x , for any x ≥ 0. This plugged into
(4.4.5), it gives the second estimate of (4.4.1), that is, there exists a constant C > 0, only
depending on a, β0, TV(u 0 ) and TV(v 0 ) such that
δ n L 1 ≤ e −β0t n /ε δ 0 L 1 + Cε.
Moreover, when ε < ∆t, we again start from the estimate (4.4.5) and note that 1/(1 −
e −β0∆t/ε ) ≤ 1/(1 −e −β0 ). Thus, there exists another constant C > 0, only depending on
a, β0, TV(u 0 ) and TV(v 0 ) such that
which gives (4.4.2).
4.4.2 Proof of Theorem 2.3
δ n L 1 ≤ e −β0t n /ε δ 0 L 1 + C∆te −β0∆t/ε ,
We are now ready to perform a rigorous asymptotic analysis of the numerical scheme
(4.2.13)-(4.2.14) when ε goes to zero.
Let us consider the numerical solution (uε h ,vε h ) to the scheme (4.2.13)-(4.2.14) written
in the form (4.3.2)-(4.3.3) with
⎧
⎨ wε h = −vε h − √ auε h ,
⎩
z ε h = +vε h − √ au ε h ,