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Publié par | profil-zyak-2012 |
Nombre de lectures | 14 |
Langue | English |
Extrait
s˙H s> 3/4
s > 5/6
3+ = 0,
Jn = 3 =u +v || 20 0 0
J
3 J( s) ˙v + v = 0 2 H
(0 ) s
2 kv k kv k0 ˙ 0 1/2H ˙H
2 2 3u+3v u+3vu +u = 0
1˙H T(ku k )˙ 10 H
1˙v H
v
1/2+2+∞ 1/2+˙ ˙v v L (H )∩L (W )t t ∞
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1˙H
2E =k∂uk2
4H =E+kuk sup E HT T4 0tT
Z ZT T
2 2 H H + v u∂ u+ vu ∂ u .T 0 t t
0 0
4L
2J(1 s)H 20