ϵ2 - Le Cermics - ENPC

24 Chapitre 2 : Analyse numérique dans un cas simplié
2.3 The local step and continuity
In each step of the domain decomposition algorithm, we must solve a quadratic
programming problem of the form (2.9). After an orthogonal change of variables,
we can assume, without loss of generality, that H is diagonal with the ordered
eigenvalues (2.10) on the diagonal and ‖a‖ = 1. Using Propositions 2.2.1 and 2.2.2,
we now determine the optimal solutions to (2.9). In the special case h = 0 and
A = a T , the rst-order optimality conditions (2.13) reduce to
Hy = yλ + aµ (2.21)
Case 1 : ɛ 1 = ɛ 2 . By Proposition 2.2.2, the multiplier λ of Proposition 2.2.1 is
λ = ɛ 1 = ɛ 2 . Dene the set
E i = {j : ɛ j = ɛ i }.
If µ ≠ 0, then by (2.21) we must have a i = 0 for all i ∈ E 1 . If i ∉ E 1 , then
y i =
µa i
ɛ i − ɛ 1
and a T y = µ ∑ i∉E 1
a 2 i
ɛ i − ɛ 1
≠ 0, (2.22)
which violates the orthogonality condition a T y = 0. Hence, µ = 0 and all y
satisfying the following conditions are solutions to (2.9) :
y i = 0 if i ∉ E 1 , a T y = 0, ‖y‖ = 1. (2.23)
Observe that there is an innite set of solutions y while the multipliers λ and
µ are unique.
Case 2 : ɛ 1 < ɛ 2 and a 1 = 0. If λ > ɛ 1 , then the second-order condition (2.14)
is violated by the vector d 1 = 1 and d i = 0 for i > 1. Hence, λ = ɛ 1 . As in Case
1, the orthogonality condition a T y = 0 is violated unless µ = 0. The solution
is again given by (2.23) and the multipliers are λ = λ 1 and µ = 0.
Case 3 : ɛ 1 < ɛ 2 and a 1 ≠ 0. We rst show that λ > ɛ 1 . Suppose, to the
contrary, that λ = ɛ 1 . The rst component of (2.13) implies that µ = 0.
Hence, (2.13) reduces to Hy = ɛ 1 y. Since H is diagonal and ɛ i > ɛ 1 for i > 1,
we conclude that y i = 0 for i > 1. Hence, y 1 = ±1 since y is a unit vector.
However, a vector of this form violates the orthogonality condition a T y = 0
when a 1 ≠ 0. This gives a contradiction, so we have λ > ɛ 1 .
(a) a i ≠ 0 for some i ∈ E 2 . We show that λ < ɛ 2 . Suppose, to the contrary,
that λ = ɛ 2 . Since a 1 ≠ 0 and a i ≠ 0 for some i ∈ E 2 , the second-order
condition (2.14) is violated by taking d to be completely zero except for
components 1 and i. Since ɛ 1 < λ < ɛ 2 , (2.21) can be solved for y :
y = µ(H − λI) −1 a. (2.24)