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

312The coefficients Y 1 , Y 2 , . . . , Y κ are computed by means of the inductiveformulas:⎧Y 1 := D 1 (Y ) − D 1 (X ) y 1 ,⎪⎨Y 2 := D 2 (Y 1 ) − D 1 (X ) y 2 ,(2.4)· · · · · · · · · · · · · · · · · · · · · · · · · · ·⎪⎩Y κ := D κ (Y κ−1 ) − D 1 (X ) y κ ,where, for 1 λ κ:(2.5) D λ := ∂∂x + y 1∂∂y + y 2∂∂y 1+ · · · + y λ∂∂y λ−1.By direct elementary computations, for κ = 1 and for κ = 2, we obtain thefollowing two very classical formulas :(2.6) ⎧⎪⎨Y 1 = Y x + [Y y − X x ] y 1 + [−X y ] (y 1 ) 2 ,Y 2 = Y x 2 + [2 Y xy − X x 2] y 1 + [Y y 2 − 2 X xy ] (y 1 ) 2 + [−X y 2] (y 1 ) 3 +⎪⎩+ [Y y − 2 X x ] y 2 + [−3 X y ] y 1 y 2 .Our main objective is to devise the general combinatorics. Thus, to attainthis aim, we have to achieve patiently formal computations of the next coefficientsY 3 , Y 4 and Y 5 . We systematically use parentheses [·] to single outevery coefficient of the polynomials Y 3 , Y 4 and Y 5 in the pure jet variablesy 1 , y 2 , y 3 , y 4 and y 5 , putting every sign inside these parentheses. We alwaysput the monomials in the pure jet variables y 1 , y 2 , y 3 , y 4 and y 5 after theparentheses. For completeness, let us provide the intermediate computationof the third coefficient Y 3 . In detail:Y 3 = D 3 (Y 2 ) − D 1 (X ) y 3( ∂=∂x + y ∂1∂y + y 2∂∂y 1+ y 3+ [Y y 2 − 2 X xy ] (y 1 ) 2 + [−X y 2](y 1 ) 3 ++ [Y y − 2 X x ]y 2 + [−3 X y ] y 1 y 2)∂∂y 2) (Y x 2 + [2 Y xy − X x 2] y 1 +(2.7)= Y x 3 1+ [2 Y x 2 y − X x 3] y 1 2+ [Y xy 2 − 2 X x 2 y] (y 1 ) 2 3 + [−X xy 2] (y 1) 3 4 ++ [Y xy − 2 X x 2]y 2 6+ [−3 X xy ] y 1 y 2 7+ [Y x 2 y]y 1 2++ [2 Y xy 2 − X x 2 y] (y 1 ) 2 3 + [Y y 3 − 2 X xy 2] (y 1) 3 4 + [−X y 3] (y 1) 4 5 +