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By Enrique Sanchez-Palencia

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H : L2(~). 2). e. a all the functions of the space Ho, Hilbert space. This is a consequence of the trace theorem, as we know. On the other hand, the Neumann condition is not imposed to all the functions of space H1, i t is satisfied only by the solution u of the abstract problem. In fact, i f we consider the subspace V~ : {v ; v E HI(~) ,. ~@ =u 0 } of H1 t h i s space is not closed (and hence is not a H i l b e r t space) and the Lax-Milgram theorem does not hold. In f a c t , the trace theorem does not work f o r - ~ n in HI .

E . , for s = 1 . . N we have I Duo +~ I = 0 ~ Duo 0 = / y @Ys dy = - ~ I Y l=m r ~ x u° : 0 Finally to obtain a well posed problem for u°, we only need a boundary condition for u°. 3) but i t w i l l be rigorously proved in next section. 11). 10). 9). 10). 2. An elementary proof, based on the maximum 5? principle is given in Bensoussan-Lions-Papanicolaou [2 ] , chap. 4. Unfortunately such a proof does not hold for other problems (such as Neumann conditions, e l l i p t i c systems . . ) . In the next section we shall give a proof due to Tartar which holds with minor modifications for many other problems.

Chap. 7) Then, f o r ~ D(~- m) we have Re ( (-~+ m)v , V)v X H : = Re [m JlvI - Re ,Vl, il - + (v 2 , Vl) V + a(v I , v2)+ ~ rlv2 il H ]= ,v2, - (v2,vl) H I > o and the operator - ~ + m is accretive. 5) or equivalently : ( ~ + ~)2 Ul + A u I = f2 + ( ~- ~ ) f I E H u2 = ( ~ + m ) Ul - f l and for ~ s u f f i c i e n t l y large we can apply the Lax-Milgram theorem to the f i r s t equation and find U l E D(A). Then, the second equation gives u2 E V. 1, we have m : 0 ( i . e . 8) of chap. 2 is s a t i s f i e d with m : 0).

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