ϵ2 - Le Cermics - ENPC

28 Chapitre 2 : Analyse numérique dans un cas simplié
By (2.32), the partial derivative ∂λ/∂a i is bounded by 2 in magnitude since both
terms in the denominator of (2.32) are positive and one of the terms is greater than
or equal to 1. Conversely, if (2.33) is violated, then we drop the rst term in the
denominator of (2.32) to obtain
∣ ∂λ ∣∣∣
∣ ≤ 4(ɛ n − ɛ 1 ) 2 .
∂a i
At this point, we have shown that there exists a constant β with the property
that if a lies on the unit sphere with a 1 ≠ 0 and a 2 ≠ 0, then
∣ ∂λ ∣∣∣
∣ ≤ β
∂a i
for each i. Observe that the solution λ to (2.25) does not change if a is multiplied
by a nonzero scalar. Hence, for any nonzero a (not necessarily on the unit sphere)
with a 1 ≠ 0 and a 2 ≠ 0, we have
∣ ∂λ ∣∣∣
∣ ≤ β/‖a‖.
∂a i
Consequently, there exists constants r 1 < 1 < r 2 with the property that
∣ ∂λ ∣∣∣
∣ ≤ 2β
∂a i
whenever r 1 ≤ ‖a‖ ≤ r 2 , a 1 ≠ 0, and a 2 ≠ 0. Given two arbitrary points on the unit
sphere, we can construct a piecewise linear path between them with the property
that the line segments all lie within the shell formed by the spheres of radius r 1 and
r 2 . Due to the bound on the partial derivatives of λ, the change in λ across each line
segment is bounded by 2β times the length of the line segment. Since the number of
line segments is bounded, independent of the location of the points, we deduce that
λ is a Lipschitz continuous function of a on the unit sphere.
Now consider the multiplier µ. If ɛ 1 = ɛ 2 , then µ = 0 (Case 1) and there is nothing
to prove. Next, we focus on Case 3a where µ is given by (2.26). For λ ∈ (ɛ 1 , ɛ 2 ), g ′
is bounded away from zero. For example,
g ′ (λ) =
n∑
i=1
a 2 i
(ɛ i − λ) ≥ 1
2 (ɛ n − ɛ 1 ) . 2
Let λ(a) denote the unique multiplier associated with any given a. Since λ is a
Lipschitz continuous function of a, it follows that µ is continuous at any point a
where λ(a) ≠ ɛ i for all i. Since λ ∈ [ɛ 1 , ɛ 2 ], the only potential points of discontinuity
are those points b for which λ(b) = ɛ i , i = 1 or i ∈ E 2 . If b 1 ≠ 0 or b i ≠ 0 for any
i ∈ E 2 , then µ(a) approaches 0 as a approaches b due to the pole in the denominator
of g ′ . Hence, µ is continuous at b.