ϵ2 - Le Cermics - ENPC

26 Chapitre 2 : Analyse numérique dans un cas simplié
where γ is the unique zero of g on the interval (ɛ 1 , ɛ 2 ). This choice for d
satises the condition a T d = 0 since g(γ) = 0. Since γ < ɛ 2 , we have
d T (H − λI)d = d T (H − ɛ 2 I)d =
n∑
d 2 i (ɛ i − ɛ 2 )
i=1
= ∑ a 2 i (ɛ i − ɛ 2 )
< ∑ a 2 i (ɛ i − γ)
= g(γ) = 0.
(ɛ i − γ) 2 (ɛ i − γ)
2
i∉E 2 i∉E 2
This violates the second-order condition (2.14).
(d) a i = 0 for all i ∈ E 2 and g(ɛ 2 ) = 0. This is the degenerate case introduced
in Section 2.1. We rst observe that λ = ɛ 2 . Suppose, to the contrary,
that λ < ɛ 2 . By (2.21), y is given by (2.24). If µ ≠ 0, then the orthogonality
condition gives (2.25), which has no solution on (ɛ 1 , ɛ 2 ) since g is monotone
and g(ɛ 2 ) = 0. Consequently, µ = 0 and (2.24) implies that y = 0, violating
the constraint y T y = 1. Thus λ = ɛ 2 .
By (2.21),
y i =
µa i
ɛ i − ɛ 2
for i ∉ E 2 . (2.28)
For i ∈ E 2 , the rst-order condition (2.13) provides no information concerning
y i since both sides of the equation vanish identically :
(ɛ i − ɛ 2 )y i = µa i = 0
The general solution is the following. First, choose any value for y i , i ∈ E 2 ,
such that
∑
i∈E 2
y 2 i ≤ 1.
Then choose µ in (2.28) such that ‖y‖ = 1. In other words, we choose µ so
that
µ 2 = 1 − ∑ i∈E 2
y 2 i
g ′ (ɛ 2 )
Notice that λ is unique in the degenerate case, while both µ and y have
multiple values.
Lemma 2.3.1 For the optimization problem (2.9) and a global minimizer y, the
multiplier λ associated with the constraint y T y = 1 is a Lipschitz continuous function
of a on the unit sphere. With appropriate sign, the corresponding multiplier µ
associated with the orthogonality constraint a T y = 0 is continuous at any nondegenerate
a.
Proof. In Case 1, Lipschitz continuity is trivially satised, so we focus on the
situation where ɛ 1 < ɛ 2 . Since the intersection of the hyperplanes a 1 = 0 or a 2 = 0
with the unit sphere are sets of measure zero on the surface of the sphere, Lipschitz
.