. Cas, Soit Ul une solution quelconque du problème (1.2) et soit 'I.t2 la solution donnée par

. Il-en-résulte-que, (E) > °tel que, pour tout, p.5

A. Pour and T. , )[, il existe 'PT E C 2 tel que: UdT,·) -E < 'PT dans D , c'est à dire, pour tout T E]O, i5(E)[ : UdT,·) < 'PT + E dans D

T. On-passe-À-la-limite-lorsque, °et puis lorsque E \ . °dans l'inégalité ci-dessus et on obtient: Ul (t) ~ M 2 (t) = U2(t) pour tout t

. Cas, Soit Ut la solution donnée par (6.18) et soit U2 une solution quelconque du problème (1.2)

=. Si-n, <. Pq, and U. Et-comme-2-<-~-on-en-déduit-que, Ti w 1 ,pq (D)) pour tout r E]I

T. Si and =. , 1 alors pq < 2(~)2 et comme r < ~;. on obtient, p.pq

I. Grâce-au-lemme and M. , pQ(D)) pour tout q 2: 1, et en particulier pour tout q > max{l, qc}, où qc = N;~~l) On peut appliquer le Principe de Comparaison, Proposition II.l3], au problème (6.47) et on a : UR(t) ~ MD,R(t) ~ M(t) pour tout t E]O, T[, UR étant définie par

. E. Donc, on peut trouver une suite ('Pn)n telle que 'Pn E C Dn et vérifie: T(to)'Pn(xO) \. B(to, xo), pp.700-706

. Alors, 27) on a : T(t)

. Comte-tenu-de, 32), par passage à la limite on obtient: limsupT(t)('Pn /\ R) ~ UR(t) dans D. n, pp.6-54

. Donc and . Pour, CPn 1\ R)(xo) :::; M(to, xo), pour tout R > O. n-+oo Alors: pour tout E > 0, (to)(CPn 1\ R)(xo) :::; M(to, xo) + E pour tout n, p.71

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