Modélisation, analyse mathématique et simulations numériques de ...

tel-00656013, version 1 - 3 Jan 2012
4.5 Proof of Theorem 2.4 117
4.5.2 Convergence proof.
Now we perform a rigorous analysis of the numerical scheme (4.2.13)-(4.2.14) when
h = (∆t,∆x) goes to zero and prove Proposition 5.1. We consider the numerical solution
(u ε h ,vε h ) to the scheme (4.2.13)-(4.2.14) and (uε ,v ε ) the exact solution to (4.1.1)-(4.1.2)
and define (w ε ,z ε ) using (4.3.1). Then we denote by
¯w n j
= 1
∆x
Cj
w ε (t n ,x)dx, ¯z n j
= 1
∆x
Cj
z ε (t n ,x)dx
and (w n j ,zn j ) (j,n)∈Z×N the numerical solution given by (4.3.2)-(4.3.3). Thus,
∆x |w n j − ¯w n j| + |z n j − ¯z n j| ≤
∆x |w n j − ˜w n j| + |z n j − ˜z n j|
j∈Z
j∈Z
+
∆x |˜w n j − ¯wn j | + |˜zn j − ¯zn j | ,
where (˜w n j ,˜zn j ) (j,n)∈Z×N is given by (4.5.4). On the one hand, we estimate the second terms
of the right hand side using the convolution properties and have
j∈Z
∆x |˜w n j − ¯w n j| + |˜z n j − ¯z n j| ≤ Cδ [TV(u) + TV(v)]. (4.5.10)
j∈Z
On the other hand, we apply the consistency error analysis to estimate the first term of
the right hand side. Applying (4.3.5)- (4.3.6) in Lemma 3.1 with (˜wj,˜zj) and (wj,zj), it
yields
and
j∈Z
j∈Z
∆x|˜w n+1
j −wn+1
∆x|˜z n+1
j −zn+1
j | ≤
j∈Z
+
j∈Z
∆x|˜w n+1/2
j
∆x|˜z n+1/2
j
+
∆x∆t |E n 1,j | + |En 2,j |
j∈Z
j | ≤
j∈Z
−
j∈Z
∆x|˜z n+1/2
j
∆x|˜w n+1/2
j
+
∆x∆t |E n 3,j | + |En 4,j | .
j∈Z
−w n+1/2
j | 1+∂wGε,∆t(wj,z n+1/2
j )
−z n+1/2
j |∂zGε,∆t(˜w n+1/2
j ,zj)
−z n+1/2
j | 1−∂zGε,∆t(˜w n+1/2
j ,zj)
−w n+1/2
j |∂wGε,∆t(wj,z n+1/2
j ).