On bifurcations from normal solutions 271
6.2. About the proof, construction of formal solutions
The starting point is the GL-system written in the form − 1
A
e
− λ
1
u = λ − λ
1
u − λu|u|
2
− 2ia · ∇ − i A
e
u − kak
2
u L a
= λ
κ
2
Im ¯u · ∇ − i Au
60 We then use the standard method. We look for a solution in the form
u = αu
1
+ α|α|
2
u
3
+ |α|
5
, a
= |α|
2
a
2
+ |α|
4
and λα
= λ
1
+ cκ |α|
2
+ |α|
4
. We can eliminate the S
1
-degeneracy by imposing α real keeping only the parity. We refer to [7] for details and just detail the beginning of the formal proof which gives the main conditions.
We first obtain, using the second equation, a
2
= λ
1
κ
2
b
2
, 61
with b
2
: = L
−1
Im ¯u
1
· ∇ − i A
e
u
1
. 62
Taking then the scalar product in L
2
with u
1
, in the first equation, we get that cκ
= λ
1
I −
2 κ
2
K ,
63 with
I :
= Z
|u
1
x |
4
d x , 64
and K
= −hib
2
· ∇ − i A
e
u
1
, u
1
i . 65
R
EMARK
9. From 63, we immediately see that there exists κ
1
such that, for κ ≥ κ
1
, cκ 0. Moreover, the uniqueness statement in Theorem 5 is true in a neighborhood which
can be chosen independently of κ ∈ [κ
1
, +∞[.
Let us now observe, that, b
2
being divergence free, it is immediate by integration by part that K
is real. Computing Re K , we immediately obtain:
K = Re K
= hL
−1
J
1
, J
1
i , 66
where J
1
is the current: J
1
: = Im ¯u
1
· ∇ − i A
e
u
1
. 67
272 M. Dutour – B. Helffer
We observe that K 0 if and only if J
1
is not identically 0. In the non simply connected case, we shall find a case when J
1
= 0. See Lemma 5. Following the argument of [7] Lemme 3.4.9, let us analyze the consequences of J
1
= 0. By assumption u
1
does not vanish identically. If u
1
x 6= 0, then we can perform in a sufficiently
small ball Bx ,
r centered at x
, the following computation in polar coordinates. We write u
1
= rx exp iθx and get J
1
= rx
2
A
e
− ∇θ = 0. So A
e
= ∇θ in this ball and this implies H
e
= 0 in the same ball. Using the properties of the zero set of u
1
in [9] and the continuity of H
e
, we then obtain H
e
= 0 in . But we know that, if is simply connected, then this implies λ
1
= 0. So we have the following lemma L
EMMA
3. If is simply connected and λ
1
0, then K 0.
Coming back to the first equation and projecting on the orthogonal space to u
1
in L
2
u
⊥ 1
, we get: u
3
= R v
3
, 68
where v
3
is orthogonal to u
1
and given by: v
3
: = 2a
2
· ∇ − i A
e
u
1
, 69
and R is the inverse of
− 1
A
e
− λ
1
on the space u
⊥ 1
and satisfies R
u
1
= 0 . We emphasize that all this construction is uniform with the parameter β
=
1 κ
in ]0, β ].
One can actually extend analytically the equation in order to have a well defined problem in [
−β , β
].
6.3. About the energy along the bifurcating solution