Ricardo Nicasso Benito A Reduç˜ao de Liapunov-Schmidt ... - Unesp

A BIFURCAÇÃO DE HOPF 63
〈v ∗ 2, v1〉 = 1
2π
= 1
2π
2π
0
2π
0
(d t 2cos(s) + d t 1sen(s))(c1cos(s) − c2sen(s))ds
[d t 2c1cos 2 (s) − d t 2c2cos(s)sen(s)
+ d t 1c1cos(s)sen(s) − d t 1c2sen 2 (s)]ds
= 1
2π
2π
0
[d t 2c1cos 2 (s) − d t 2c1sen 2 (s)
− d t 1c1cos(s)sen(s) + d t 1c1cos(s)sen(s)]ds
= 1
2π
= 1
2π
2π
0
2π
Para o caso (c) temos:
< v1, v1 > = 1
2π
= 1
2π
0
2π
0
2π
0
[d t 2c1(cos 2 (s) − sen 2 (s))]ds
0ds = 0.
(c t 1cos(s) − c t 2sen(s))(c1cos(s) − c2sen(s))ds
[c t 1c1cos 2 (s) − c t 1c2cos(s)sen(s)
− c t 2c1cos(s)sen(s) + c t 2c2sen 2 (s)]ds
= 1
2π
2π
0
[c t 1c1cos 2 (s) + (2 − c t 1c1)sen 2 (s)
− (c t 1c2 + c t 2c1)cos(s)sen(s)]ds
= 1
2π
2π
0
[c t 1c1cos 2 (s) + 2sen 2 (s) − c t 1c1sen 2 (s)
− 2c t 1c2cos(s)sen(s)]ds
= 1
2π
= 1
2π
−c t 1c2
= 1
2π
= 1
2π
2π
0
c t 1c1
2π
0
[c t 1c1cos(2s) + 2sen 2 (s) − c t 1c2sen(2s)]ds
2π
cos(2s)ds + 2
0
sen(2s)ds
2π
c t sen(2s)
1c1 |
2
2π
0 + s| 2π
0 − sen(2s)
2
2π + ct1c2 2 − ct
1c2
= 1.
2
0
(1 − cos(2s))ds
| 2π
0 + c t 1c2
cos(2s)
|
2
2π
0