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

2768.52. Preparation. Thus, translating the considerations to the PDE language,with n = 2 and m = 1, consider a submanifold of solutions ofthe form(8.53)y = b + Π(x, a)= b + 2 x1 a 1 + x 1 x 1 a 2 + a 1 a 1 x 21 − x 2 a 2 + O 4 ,where O 4 is a function of (x, a) only. The term 2 x 1 a 1 corresponds to a Leviform of rank 1 at every point. The term x 1 x 1 a 2 guarantees solvabilitywith respect to the parameters (compare Definition 2.12). Let us develope(8.54) Π(x, a) = ∑ ∑Π k1 ,k 2 ,l 1 ,l 2(x 1 ) k 1(x 2 ) k 2(a 1 ) l 1(a 2 ) l 2,k 1 ,k 2 0 l 1 ,l 2 0with Π k1 ,k 2 ,l 1 ,l 2∈ K. Of course, Π 1,0,1,0 = 2, Π 2,0,0,1 = 1 and Π 0,1,2,0 = 1.Lemma 8.55. A transformation belonging to G v,p insures(8.56)Π k1 ,k 2 ,0,0 = 0, k 1 + k 2 0, Π 0,0,l1 ,l 2= 0, l 1 + l 2 0,Π k1 ,k 2 ,1,0 = 0, k 1 + k 2 2, Π 1,0,l1 ,l 2= 0, l 1 + l 2 2,Π k1 ,k 2 ,2,0 = 0, k 1 + k 2 2, Π 2,0,l1 ,l 2= 0, l 1 + l 2 2.Proof. Lemma 7.11 achieves the first line. The monomial x 1 being factoredby [a 1 +O 2 (a)], we set a 1 := a 1 +O 2 (a) to achieve Π 1,0,l1 ,l 2= 0, l 1 +l 2 2.As in the proof of Lemma 7.18, we pass to the dual equation b = y −Π(x, a)to complete Π k1 ,k 2 ,1,0 = 0, k 1 + k 2 2. Finally, x 1 x 1 is factored by [a 2 +O 2 (a)], so we proceed similarly to achieve the third line.Since Π(x, a) is assumed to be independent of b, the assumption that theLevi form of M ⊂ C 3 has exactly rank 1 at every point translates to:(8.57) 0 ≡∣ Π x 1 a 1 Π ∣x 1 a ∣∣∣ 2Π x 2 a 1 Π .x 2 a 2For later use, it is convenient to develope somehow Π with respect to thepowers of (a 1 , a 2 ):(8.58)y = b + 2 x1 a 1 + x 1 x 1 a 2 + a 1 a 1 x 21 − x 2 a 2 ++ a 2 b(x) + a 1 a 2 d(x) + a 2 a 2 e(x) + a 1 a 1 a 1 f(x) + a 1 a 1 a 2 g(x)++ (a 1 ) 4 R + (a 1 ) 3 a 2 R + a 1 (a 2 ) 2 R + (a 2 ) 3 R,with R = R(x, a) being an unspecified remainder. Thanks to the previouslemma, the coefficients a of a 1 and c of a 1 a 1 must vanish. The function b isan O 3 .