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Additional info for Solutions to the nonlinear Schrodinger equation carrying momentum along a curve. I study of the limit set and approximate solutions
Rn−1 Rn−1 U (kz)p−2 ∂m U (kz)wr2 = 1 hk wr ∂m F. Rn−1 Hence A4 can be written as A4 = − + 1 k 2 H l ∂ml U (kz) + (f )2 ∂m U (kz) wr k Rn−1 l H l (zl + Φl ) l 2 wr − (f ) U (kz)H m N − k∂m U (kz) ∇ V, z + Φ − hU (kz) ∇N V, Em Rn−1 ∂m U (kz) wr ∇N V, z + Φ + 2(f )2 H, z + Φ wr + Rn−1 + h2 k 2 H l ∂l wr . ∂m U (kz) Rn−1 l 32 After some cancelations and some integration by parts we find the following formula Hl A4 = −2k Rn−1 l 2 wr ∂ml U (kz) + p−1 2 k 2kθ wr U (kz). Rn−1 Using the symmetries of the integrals we find 1 n−1 = −2kH m A4 = −2kH m p−1 wr U (kz) 2θ Rn−1 1 p−1 1 + − U p (kz) .
5. 37 Proof of (35). 5, taking into account (19) and (20) we obtain (66) Σ1 + Σ2 = 0, where Σ1 A2 σhσ−1 = γ ∂h N ∇ V, V ds + ∂V Aσ(σ − 1)hσ−2 γ ∂h N ∂h ∂h N ∇ V, W + A ∇ V, V ds ∂V ∂A 2 ∂V ∂2h ∂2h N ∇ V, W + + Aσhσ−1 A ∇N V, V ds + ∂V 2 ∂V ∂A 2 γ ∂h N − A2 hσ H, V ds − Aσhσ−1 ∇ V, V H, W ds ∂V γ γ − ∂h ∂h N ∇ V, W + A ∂V ∂A 2 hσ−1 Aσ γ ˙j− V˙ j W Ahσ = γ R1j1m V j W m ds + A1 A2 σ j γ ∂h ds + AA1 σ(σ − 1) ∂A 2 γ 2 ∂h N ∂h ∇ V, W + A ds − A1 ∂V ∂A 2 hσ−1 γ ∂h ds ∂A ∂h N ∂h ∂h ∇ V, W + A2 ds ∂V ∂A ∂A hσ−2 ∂ h ∂ h ∇N V, W + A ds − AA1 σ ∂A∂V ∂A2 2 hσ−1 γ + A1 σ hσ−1 γ j,m hσ−1 + AA12 σ + AA1 σ γ ∂h (∇N )2 V [V, W]ds ∂V H, V ds; Σ2 Aσhσ−1 hσ−1 γ ∂h H, W ds ∂A hσ H, W ds + A12 γ hσ ds.
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