Modélisation, analyse mathématique et simulations numériques de ...

tel-00656013, version 1 - 3 Jan 2012
4.5 Proof of Theorem 2.4 115
On the other hand, we proceed to the evaluation of the terms En 23,j , En 24,j and En 25,j . First,
for s ∈ [0,∆t], we set
ϕδ,x(s) = [R(u,v)⋆ρδ](t n +s,x− √ a(∆t−s)).
Then, from (4.2.2) and (4.2.6), we know that |∂uR(u,v)| ≤ √ aβ and |∂vR(u,v)| ≤ β, for
any (u,v) ∈ I(N0,a0), we obtain
j∈Z
∆x∆t
n
E
1
23,j ≤
ε
R
≤ C ∆t
ε
+ C ∆t
ε
∆t
s
ϕ
0 0
′ δ,x
tn+1
R tn tn+1 R
t n
(η)dη ds
dx, (|∂tuδ| + |∂xuδ|)(t,x)dt dx
(|∂tvδ| + |∂xvδ|)(t,x)dt dx.
Thus we can use the estimates on the continuous relaxation system listed in Theorem 1.1.
Indeed, since
⎧
⎪⎨
∂tuδ = −∂xvδ,
⎪⎩
∂tvδ = −a∂xuδ − 1
ε Rδ(u,v),
we obtain, by applying a first order Taylor expansion of R, the inequalities
R
R
(|∂tuδ| + |∂xuδ|)(t,x)dx ≤ TV(u(t)) + TV(v(t)),
(|∂tvδ| + |∂xvδ|)(t,x)dx ≤ C TV(u(t))+ 1
ε (v −A(u))(t)L1
.
Hence, integrating over t ∈ (t n ,t n+1 ) and using (4.1.6) and (4.1.7), we get:
j∈Z
∆x E n
23,j ≤ C ∆t
ε
TV(u(t n )) + TV(v(t n )) + e−β0t n /ε
δ
ε
0
L1 + 1 , (4.5.7)
where C > 0 only depends on √ a and β.
Now we treat the term En 24,j using the smoothness properties of R (4.2.2) and (4.2.6),
it gives
j∈Z
∆x|E n 24,j| = 1
ε
≤ C
ε
R
R
n √ n √
R(u,v)(t ,x−y − a∆t)−R(u,v)(t ,x− a∆t) ρδ(y)dy
dx,
R2 [|u(t n ,x)−u(t n ,x−y)| + |v(t n ,x)−v(t n ,x−y)|]ρδ(y)dy dx.