ϵ2 - Le Cermics - ENPC

34 Chapitre 2 : Analyse numérique dans un cas simplié
Combining (2.47)(2.50), it follows that when k is large enough that |F j(0)| ′ ≤ η
for all j ≥ k, we have
µ k−1,1 ≽ µ k1 ≽ µ k2 ≽ µ k+1,2 ≽ µ k+1,1 (2.51)
where the notation µ k1 ≽ µ k2 means that µ k2 ≤ µ k1 + η. Hence, in each iteration,
the µ multiplier either decreases or makes an increase which is bounded by η.
Step 3. The case lim inf µ k1 = 0.
When lim inf µ k1 = 0, there exists a subsequence of the iterates with the property
that µ k1 tends to 0. By (2.51) and the fact that η can be taken arbitrarily small, we
conclude that the corresponding subsequence of the multipliers µ k2 also approaches
0. Since y k lies in a compact set, we can extract subsequences converging to a limit
denote y. By (2.44), the corresponding subsequence of multipliers λ k1 and λ k2 also
approach limits denoted λ 1 and λ 2 . By (2.43), we have
[
H1 0
0 H 2
] [
y1
y 2
]
=
[ ]
λ1 y 1
.
λ 2 y 2
Since y 1 and y 2 are orthogonal unit vectors, we conclude that y is a stationary point
for (2.1) corresponding to the multiplier µ = 0.
Step 4. The case µ = lim inf µ k1 > 0.
Let us focus on a subsequence of the iterates, also denoted µ k1 for convenience,
with the property that µ k1 approaches µ. Given any η > 0, choose K large enough
that (2.51) holds for all k ≥ K. Also, choose K large enough that
Hence, for some k ≥ K, we have
∣ µ − inf
k≥K µ k1
∣
µ − η ≤ µ k1 ≤ µ + η.
By (2.51) it follows that
µ k2 ≤ µ k1 + η ≤ µ + 2η.
Also, by (2.51) and by (2.52), we have
Combining these inequalities gives
µ − η ≤ µ k+1,1 ≤ µ k+1,2 + η ≤ µ k2 + 2η.
∣ ≤ η 2 . (2.52)
µ − 3η ≤ µ k2 ≤ µ + 2η and µ − η ≤ µ k1 ≤ µ + η.
Since η is arbitrary, it follows that µ k1 and µ k2 approach the same limit µ.
Again, by extracting subsequences, there exist limits y 1 , y 2 , λ 1 , and λ 2 such that
[ ]
λ1 y 1 + µx 2
. (2.53)
µy 1 + λ 2 y 2
[
H1 0
0 H 2
] [
y1
y 2
]
=