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

A BIFURCAÇÃO DE HOPF 61
(i) d t c = 2, isto é:
(ii) d t c = 0, isto é:
Logo, de (4.40) e (4.41) temos que:
Com isso temos:
〈v ∗ 1, v1〉 = 1
2π
= 1
2π
2π
0
2π
0
d t 1c1 + d t 2c2 = 2
d t 1c2 − d t 2c1 = 0
d t 1c1 − d t 2c2 = 0
d t 1c2 + d t 2c1 = 0
d t 1c2 = d t 2c1 = 0
d t 1c1 = d t 2c2 = 1
(d t 1cos(s) − d t 2sen(s))(c1cos(s) − c2sen(s))ds
[d t 1c1cos 2 (s) − d t 1c2cos(s)sen(s)
− d t 2c1cos(s)sen(s) + d t 2c2sen 2 (s)]ds
= 1
2π
2π
0
[d t 2c2cos 2 (s) + d t 2c2sen 2 (s) − (d t 1c2
+ d t 2c1)cos(s)sen(s)]ds
= 1
2π
= 1
2π
2π
0
2π
0
= 1
(2π) = 1.
2π
[d t 2c2(cos 2 (s) + sen 2 (s)) − 0]ds
ds = 1
2π (s) 2π
0
(4.40)
(4.41)