158
(11.2,7)
(u, Curlq) - (curlu, q) = < a (u )n, q >
Using (11.2,5) and (11.2,7) we get the variational formulation associated with the
problem (PI) :
To find u belonging to V and A belonging to R :
(11.2,8)
(divu,divv) + (curlu, curlv) = A (u,v) and for each v E V
where the space V that was defmed in (2,1) is a Hilbert's space for the norm:
(11.2,9)
lIu 11= ( 1 divu 12 + 1 curlu 12 + 1 u 12) 112
We note that if u E V then u E H I loc ' but if Q is sufficiently smooth then we have u
E H1(Q)2.
This result is given in the
II.2.3 Lemma 1 [Girault - Raviart (1986)]
If Q is an open subset of R 2 , of boundary Y of class C 1 , 1 where ' Y is a convex polygon,
then V is algebraically and topologicaly included into Hl( Q)2:
(11.2,10)
and the norm (2,9) is on V equivalent to the H 1( Q)2 norm'
Besides, we have the orthogonal decomposition in L 2 (Q)2 and in V:
(11.2,11)
u = gradp + Curlq
with p defined at one close constant, solution of :
(11.2,12)
-Ap = divu into Q; gradp. n = 0 on y
and q solution of :
(11.2,13)
-Aq = curlu into Q ; q = 0 on Yo' q = ci on Yi
where the ci are constants determined by (11.2,11) and (11.2,12).
Note : the regularity conditions given above come from regularity results obtained by P.
Grisvard (1981): the solutions of the problems (11.2,12) and (11.2,13) are in H2 (Q) if the
conditions (IT. 1 , 1) are realised, and therefore gradp and Curlq are in HI (Q) 2 .
In the following corollary we give a decompostion of the space V which is better
adapted to the demonstration of the theorem.