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

and moreover, all other, higher order partial derivatives of X , of Y 1 and ofY 2 may be expressed as P ( X , Y 1 , Y 2 , Y 1x , X x).Corollary 8.20. An infinitesimal Lie symmetry of (E 4 ) is uniquely determinedby the five initial Taylor coefficients271(8.21) X (0), Y 1 (0), Y 2 (0), Y 1x (0), X x(0).Proof of the proposition. We notice that (8.18) 1 and (8.18) 3 are given forfree by (8.13) 1 and by (8.15) 1 . Differentiating (8.13) 3 with respect to x, weget:(8.22)0 = −Y xy 2 − X xx + 2 Y 1xy + (6 + 1 3g1 x)Y 1y + (6x + 2 3g1 )Y 1xy + r Y 1 +2+ r Yx 1 + r Y 2 + r Yx 2 + r Y x 1 + s Y xx 1 + r X + g2 x [1 + g2 ] −1 X x .By (8.15) 1 , s Yxx 1 vanishes. We replace Yx 2 thanks to (8.13) 1 . Differentiating(8.13) 1 with respect to y 2 , we may substract 0 = −Y xy 2 + (2x + g 1 )Y 1xy+ 2r Yx 1 . We get:(8.23)0 = −X xx + 2 Y 1xy + (4x + 1 2g1 )Y 1xy 2++ (6 + 3gx)Y 1 1y + r X + r Y 1 + r Y 2 + r Y 12 x + gx[1 2 + g 2 ] −1 X x .By means of (8.15) 2 , we replace the first three terms and then solve Y 1y 2 :(8.24) Y 1y 2 = r X + r Y 1 + r Y 2 + r Y 1x + k ∗ X x ,introducing a notation for a new function that should be recorded:(8.25) k ∗ := g 2 x [1 + g2 ] −1 [4 + 3g 1 x − h]−1 .This is (8.18) 10 . Next, we differentiate the obtained equation with respect tox, getting:(8.26) Y 1xy 2 = r X + r Y 1 + r Y 2 + r Y 1x + r X x + k ∗ X xx .This is (8.18) 13 . We replace the obtained value of Y 1yin (8.13) 2 2 , (8.13) 3 ,(8.15) 2 and the obtained value of Y 1xyin (8.15) 2 2 . This yields a new, simpler