About the proof, construction of formal solutions

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