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

Lemma 8.59. The function b depends only on x 1 , is an O 3 (x 1 ) and thefunction g satisfies g x 2 x2(0) = 0.Proof. Developing [1 −x 2 a 2 ] −1 = 1+x 2 a 2 +(x 2 a 2 ) 2 +O 3 (x 2 a 2 ), insertingthe right hand side of(8.60)y − b = a 1[ 2x 1] + a 2[ x 1 x 1 + b(x) ] + a 1 a 1[ x 2] + a 1 a 2[ 2x 1 x 2 + d(x) ] +277+ a 2 a 2[ x 1 x 1 x 2 + e(x) ] + a 1 a 1 a 1[ f(x) ] + a 1 a 1 a 2[ x 2 x 2 + g(x) ] ++ (a 1 ) 4 R + (a 1 ) 3 a 2 R + a 1 (a 2 ) 2 R + (a 2 ) 3 Rin the determinant (8.57) and selecting the coefficients of cst., of a 1 , of a 2and of a 1 a 1 , we get four PDEs:(8.61)0 = 2b x 2,0 = 2d x 2 − 2b x 1,0 = 4e x 2 − 2x 1 d x 2 − 2x 1 b x 1 − d x 2 b x 1,0 = 2g x 2 − 2d x 1 − [ 6x 1 + 3b x 1]fx 2.The first one yields b = b(x 1 ), which must be an O 3 (x 1 ), because thewhole remainder is an O 4 . Differentiating the fourth with respect to x 2 , itthen follows that g x 2 x2(0) = 0.8.62. Associated PDE system (E 5 ). Next, differentiating (8.60) with respectto x 1 , to x 1 x 1 and to x 1 x 1 x 1 , we compute y x 1 and y x 1 x 1, we substitute y 1 andy 1,1 and we push the monomials a 2 a 2 , a 1 a 1 a 1 and a 1 a 1 a 2 in the remainder:(8.63)y 1 = 2a 1 + a 2 [2x 1 + b x 1] + a 1 a 2 [2x 2 + d x 1] + (a 2 ) 2 R + (a 1 ) 3 R + (a 1 ) 2 a 2 R,y 1,1 = a 2 [2 + b x 1 x 1] + a1 a 2 [d x 1 x 1] + (a2 ) 2 R + (a 1 ) 3 R + (a 1 ) 2 a 2 R,y 1,1,1 = a 2 [b x 1 x 1 x 1] + a1 a 2 [d x 1 x 1 x 1] + (a2 ) 2 R + (a 1 ) 3 R + (a 1 ) 2 a 2 R.Here, the written remainder cannot incorporate a 1 a 1 , so it is said that thecoefficient of a 1 a 1 does vanish in each equation above. Solving for a 1 anda 2 from the first two equations, we get(8.64)⎧a 1 = 1 [ 2x 1 ] [⎪⎨ 2 y + b1 − y x 1 2x 2 ]+ d1,1 − y 1 y x 11,1 +4 + 2b x 1 x 1 8 + 4b x 1 x 1+ (y 1,1 ) 2 R + (y 1 ) 3 R + (y 1 ) 2 y 1,1 R,[ ] [ ]⎪⎩ a 2 1d= y 1,1 − y 1 y x 1 x 11,1 + (y 1,1 ) 2 R + (y 1 ) 3 R + (y 1 ) 2 y 1,1 R.2 + b x 1 x 1 2(2 + b x 1 x 1)2