Travaux sur les symétries de Lie des équations aux ... - DMA - Ens

288where a, b, c, d, e, f, g, h, j, k ∈ K are arbitrary. Computing the third prolongationsof the ten vector fields(8.120)∂∂x 1,∂ ∂∂x 2, ∂y ,− x 2 ∂∂x 1 + ∂ 2x1 ∂y , ∂x1− x 1 x 2 ∂∂x 1 − x2 x 2x 1 y ∂∂x 1 − x1 x 1 ∂∂x 2 + yy ∂ ∂y∂x 1 + 2y ∂ ∂y , ∂x1∂x 1 + 2x2∂∂x 2 + x1 x 1 ∂ ∂y ,∂∂x 2,−y ∂∂x 1 + ∂2x1 ∂x 2,(x1 x 1 − x 2 y) ∂∂x 1 + 2x1 x 2 ∂∂x 2 + 2x1 y ∂ ∂y ,one verifies that they all are tangent to the skeleton y 2 = 1 4 (y 1) 2 , y 1,1,1 = 0.Thus the bound is attained. One then verifies ([FK2005a]) that the spannedLie algebra is isomorphic to so(5, C).Lemma 8.121. Assuming the normalizations of Lemma 8.54, the remainderO 4 in (8.53) is an O 3 (x 1 , a 1 ):(8.122)y = b+ 2 x1 a 1 + x 1 x 1 a 2 + a 1 a 1 x 21 − x 2 a 2 +(x 1 ) 3 R+(x 1 ) 2 a 1 R+x 1 (a 1 ) 2 R+(a 1 ) 3 R.Proof. Indeed, writing(8.123)y = b + x 1 Λ 1,0 + a 1 Λ 0,1 + x 1 x 1 Λ 2,0 + x 1 a 1 Λ 1,1 + a 1 a 1 Λ 0,2 + O 3 (x 1 , a 1 ),with Λ i,j = Λ i,j (x 2 , a 2 ), and developing the determinant (8.54) with respectto the powers of (x 1 , a 1 ), the vanishing of the coefficients of cst., of x 1 , ofa 1 yields the system(8.124)⎧⎪⎨⎪⎩0 ≡ Λ 1,0a 2 Λ 0,1x 2 ,0 ≡ Λ 1,1 Λ 1,0x 2 a 2 − 2 Λ 2,0a 2 Λ 0,1x 2 − Λ 1,1x 2 Λ 1,0a 2 ,0 ≡ Λ 1,1 Λ 0,1x 2 a 2 − Λ 1,1a 2 Λ 0,1x 2 − 2 Λ 0,2x 2 Λ 1,0a 2 .If the first equation yields Λ 1,0a≡ 0, replacing in the second, using Λ 2,0 =2a 2 +O 2 , we deduce that Λ 0,1x≡ 0 also. Similarly, Λ 0,12 x≡ 0 implies Λ 1,02 a≡ 0. 2Since the coordinate system satisfies the normalization Π(0, a) ≡ Π(x, 0) ≡0, necessarily Λ 1,0 = O(a 2 ) and Λ 0,1 = O(x 2 ). We deduce:(8.125) 0 ≡ Λ 1,0 ≡ Λ 0,1 .