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

260(Denote by ∆(x, a, b) the determinant of the (invertible) matrixΠj(x, a, b) ) and by D(x, a, b) its matrix of cofactors, so thatb l 1l,jmΠ −1b= [∆] −1 D. Hence we can solve G from (6.43):(6.44) ⎧[D(x, a, b)G (a, b) ≡ Y ( x, Π(x, a, b) ) n∑− X ⎪⎨i( x, Π(x, a, b) ) Π x i(x, a, b)−∆(x, a, b)i=1]p∑− F ⎪⎩q (a, b) Π a q(x, a, b) .q=1Next, we aim to solve the ( F q (a, b). Consequently, we gather all the otherterms in the brackets as Ψ 0 J1x,a,b Π, X , Y ) :(6.45)G (a, b) ≡[D(x, a, b)−∆(x, a, b)]p∑F q (a, b) Π a q(x, a, b)q=1+ Ψ (0 J1x,a,b Π, X , Y ).∆(x, a, b)Here, Ψ 0 is linear with respect to (X , Y ), with polynomial coefficients ofdegree one in Jx,a,b 1 Π.Next, for k = 1, . . ., n, we differentiate this identity with respect to x k .Then G (a, b) disappears and we chase the denominator ∆ 2 :(6.46) ⎧ []p∑0 ≡ [∆ D] − F q (a, b) Π a q x k(x, a, b) +⎪⎨q=1[]p∑+ [∆ D xk − ∆ xk D] − F q (a, b) Π a q(x, a, b) +⎪⎩q=1+ Ψ k(J2x,a,b Π, J 1 x,y X , J1 x,y Y ) .The Ψ k are linear with respect to (Jx,y 1 X , J1 x,yY ), with polynomial coefficientsin Jx,a,b 2 Π. Then we further differentiate with respect to x and byinduction, for every β ∈ N n , we get:(6.47) ⎧ []p∑0 ≡ [∆ D] − F q (a, b) Π a q x β(x, a, b) +q=1⎪⎨+ ∑ (D β,β1 J|β 1 |+1 Π )[ ]p∑− F q (a, b) Π a q x 1(x, a, b) +β⎪⎩|β 1 |