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

268• r and s are unspecified functions, varying in the context, of (x, y 1 , y 2 )with s = O(x) + O(y 1 ) + O(y 2 ), but possibly r(0) ≠ 0;• R is a remainder function of all the variables (x, y 1 , y 2 , y 1 1) parametrizing∆ E4 .Letting L = X ∂ + Y 1 ∂ + Y 2 ∂ be a candidate infinitesimal Lie∂x ∂y 1 ∂y 2symmetry and applying L (2) = L + Y1 1 ∂ + Y 2 ∂y11 1 ∂ + Y 1 ∂y12 2 ∂ + Y 2 ∂y21 2 ∂∂y22to ∆ E4 , we obtain firstly, computing mod (y1) 1 5 :(8.7)0 ≡ −Y1 2 + [ X ]( y1(2 1 + gx) 1 + (y1) 1 2 gx 2 + (y1) 1 3 r + (y1) 1 4 r ) ++ [ Y 1]( y1 1 r + (y1) 1 2 r + (y1) 1 3 r + (y1) 1 4 r ) ++ [ Y 2]( y1 1 r + (y1) 1 2 r + (y1) 1 3 r + (y1) 1 4 r ) ++ Y1( 1 2x + g 1 + y1 1 (2 + 2g2 ) + (y1 1 )2 s + (y1 1 )3 s + (y1 1 )4 s ) ,and secondly, computing mod (y 1 1 )2 :(8.8) 0 ≡ −Y 1 2 + 2 y1 1 Y1 1 h.The third Lie equation involving Y2 2 will be superfluous. Specializing(4.6)(II) to m = 2, we get Y1 1 and Y1:2(8.9)Y1 1 = Y x 1 + [ Y 1y − X ] 1 x y11 + [ ]Y 1y y2 2 1 + [ − X y 1](y11 ) 2 + [ − X y 2]y11 y12Y1 2 = Y x 2 + [ ]Y 2y y1 1 1 + [ Y 2y − X x]y2 2 1 + [ − X y 1]y21 y1 1 + [ − X y 2](y21 ) 2 .and also Y2 1 and Y2 2 (in fact superfluous):(8.10)Y2 1 = Y xx 1 + [ 2 Y 1xy − X ] 1 xx y11 + [ ]2 Y 1xy y2 2 1 + [ Y 1y 1 y − 2 X ] 1 xy 1 (y11 ) 2 ++ [ 2 Y 1y 1 y − 2 X ] 2 xy 2 y11 y1 2 + [ ]Y 1y 2 y (y2 2 1 ) 2 + [ ]− X y 1 y 1 (y11 ) 3 ++ [ ]− 2 X y 1 y 2 (y11 ) 2 y1 2 + [ ]− X y 2 y 2 y11 (y1 2 )2 + [ Y 1y − 2 X x]y1 1 2 ++ [ ]Y 1y y2 2 2 + [ − 3 X y 1]y11 y2 1 + [ − X y 2]y11 y2 2 + [ − 2 X y 2]y21 y2 1 ,Y2 2 = Yxx 2 + [ ]2 Y 2xy y1 1 1 + [ 2 Y 2xy − X ] 2 xx y21 + [ ]Y 2y 1 y (y1 1 1 ) 2 ++ [ 2 Y 2y 1 y − 2 X ] 2 xy 1 y11 y1 2 + [ Y 2y 2 y − 2 X ] 2 xy 2 (y21 ) 2 + [ ]− X y 1 y 1 (y11 ) 2 y1+2+ [ ]− 2 X y 1 y 2 y11 (y1) 2 2 + [ ]− X y 2 y 2 (y21 ) 3 + [ ]Yy 2 y1 1 2 ++ [ Y 2y − 2 X x]y2 2 2 + [ − 2 X y 1]y11 y2 2 + [ − X y 1]y21 y2 1 + [ − 3 X y 2]y21 y2.2Inserting Y1 2 and Y1 1 in the first Lie equation (8.7) in which y2 1 is replaced bythe value (8.6) 1 it has on ∆ E4 and still computing mod(y1 1)5 , we get, again