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

250coefficient(4.15) Ŷ j i 1 ,...,i λ= Ŷj i 1 ,...,i λ(xi 1, y j 1, y j(q 1)β(q 1 ) , Jλ x,yX i 1, J λ x,yY j 1 ) ,that depends linearly on the λ-th jet of the coefficients of L , as confirmedby an inspection of Part II’s formulas. Here, we use the jet notationJx,y λ Z := ( ∂ α 1x ∂β 1y Z ) |α 1 |+|β 1. We thus get equations|λn∑(4.16) 0 ≡ −Ŷj α + X i ∂F αj n∑+ Y l ∂F j p∑α+ Ŷ j(q) ∂Fαj∂x i ∂y l β(q),i=1l=1q=1∂y j(q)β(q)involving only the variables ( x i 1, y j 1, y j(q 1)β(q 1 )).Next, we develope every such equation with respect to the powers ofy j(q 1)β(q 1 ) :(4.17)0 ≡ ∑µ 1 ,...,µ p0(y j(1)β(1) )µ1 · · ·(y j(p)β(p) )µp Ψ j α,µ 1 ,...,µ p(xi 1, y j 1, J κ+1x,y X i 1, J κ+1x,y Y j 1 ) .The Ψ j α,µ 1 ,...,µ pare linear with respect to ( J κ+1x,y X i 1, J κ+1x,y Y j 1), with certaincoefficients analytic with respect to (x, y), which depend intrinsically (butin a complex manner) on the right hand sides F j α.Proposition 4.18. The vector field L is an infinitesimal Lie symmetry of(E ) if and only if its coefficients X i 1, Y j 1satisfy the linear PDE system:((4.19) 0 = Ψ j α,µ 1 ,...,µ p xi 1, y j 1, Jx,y κ+1 X i 1, Jx,y κ+1 Y ) j 1for all (j, α) ≠ (j, 0) and ≠ (j(q), β(q)) and for all (µ 1 , . . .,µ p ) ∈ N p .In all known instances, a finite number of these equations suffices.Example 4.20. With n = m = κ = 1, a second prolongation L (2) =X ∂ + Y ∂ + Y ∂x ∂y 1 ∂ ∂∂y 1+ Y 2 ∂y 2is tangent to the skeleton 0 = −y 2 +F(x, y, y 1 ) of (E 1 ) if and only if 0 = −Y 2 + X F x + Y F y + Y 1 F y1 , or,developing:(4.21) ⎧⎪⎨0 = −Y xx + [ ]− 2 Y xy + X xx y1 + [ ]− Y yy + 2 X xy (y1 ) 2 + [ ]X yy (y1 ) 3 ++ [ ] [ ]− Y y + 2 X x F + 3 Xy y1 F + [ X ] F x + [ Y ] F y +⎪⎩+ [ ]Y x Fy1 + [ ]Y y − X x y1 F y1 + [ ]− X y (y1 ) 2 F y1 .Developing F = ∑ k0 (y 1) k F k (x, y), we may obtain equations (4.19).§5. EXAMPLES5.1. Second order ordinary differential equation. Pursuing the study of(E 1 ), according to Section 7 below, we may assume that F = O(y x ), orequivalently F(x, y, 0) ≡ 0.