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

278We then get (notice the change of remainder):(8.65)a 1 a 1 = 1 [ 2x4 (y 1) 2 1 ] [+ b− y 1 y x 12x1,1 − (y 1 ) 2 2 ]+ dy x 11,1 + (y 1 ) 3 R + (y 1,1 ) 2 R,4 + 2b x 1 x 1 8 + 4b x 1 x[ ] [ ]1a 1 a 2 1= y 1 y 1,1 − (y 1 ) 2 dy x 1 x 11,1 + (y 1 ) 3 R + (y 1,1 ) 2 R,4 + 2b x 1 x 1 (4 + 2b x 1 x 1)2a 2 a 2 = (y 1 ) 3 R + (y 1,1 ) 2 R,[ 6xa 1 a 1 a 1 = −(y 1 ) 2 1 ]+ 3by x 11,1 (y 1 ) 3 R + (y 1,1 ) 2 R,16 + 8b x 1 x[ ]1a 1 a 1 a 2 = (y 1 ) 2 1y 1,1 + (y 1 ) 3 R + (y 1,1 ) 2 R.8 + 4b x 1 x 1Differentiating (8.60) with respect to x 2 , substituting y 2 for y x 2 and replacingd x 2 by b x 1 thanks to (8.61) 2 , we get(8.66)y 2 = a 1 a 1 + a 1 a 2 [2x 1 + b x 1] + a 2 a 2 [x 1 x 1 + e x 2] + a 1 a 1 a 1 [f x 2] + a 1 a 1 a 2 [2x 2 + g x 2]++ (a 1 ) 4 R + (a 1 ) 3 a 2 R + a 1 (a 2 ) 2 R + (a 2 ) 3 R.Replacing the monomials (8.65), we finally obtain:(8.67)y 2 = 1 [ ]4 (y 1) 2 + (y 1 ) 2 2gx 2 − 2d x 1 − (6x 1 + 3b x 1)f x 2y 1,1 − (2x1 + b x 1)d x 1 x 1+16 + 8b x 1 x 1 (4 + 2b x 1 x 1)2+ (y 1 ) 3 R + (y 1,1 ) 2 R.Thanks to (8.61) 4 , the first (big) coefficient of (y 1 ) 2 y 1,1 is zero; then theremainder coefficient is an O(x 1 ), hence vanishes at x = 0, together with itspartial first derivative with respect to x 2 . Accordingly, by s ∗ = s ∗ (x 1 , x 2 ),we will denote an unspecified function satisfying(8.68) s ∗ (0) = 0 and s ∗ x 2(0) = 0 .Lemma 8.69. The skeleton of the PDE system (E 5 ) associated to the submanifold(8.58) possesses three main equations of the form(∆ ⎧ E5 )y 2 = 1 4 (y 1) 2 + (y 1 ) 3 r + (y 1 ) 4 r + (y 1 ) 5 r + (y 1 ) 6 R+⎪⎨+ y 1,1[(y1 ) 2 s ∗ + (y 1 ) 3 r + (y 1 ) 4 r + (y 1 ) 5 r ] + (y 1,1 ) 2 R,y 1,2 = 1 2 y 1y 1,1 + (y 1 ) 3 r + (y 1 ) 4 r + (y 1 ) 5 r + (y 1 ) 6 R+[+ y 1,1 (y1 ) 2 r + (y 1 ) 3 r + (y 1 ) 4 r + (y 1 ) 5 r ] + (y 1 ) 6 R,[y 1,1,1 = (y 1 ) 3 r + (y 1 ) 4 r + y 1,1 r + y1 r + (y 1 ) 2 r + (y 1 ) 3 r ] +⎪⎩+ (y 1,1 ) 2[ r + y 1 r + (y 1 ) 2 r + (y 1 ) 3 r ] + (y 1,1 ) 3 R,