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

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Observe that these expressions are polynomial in the jet variab<strong>les</strong>, their coefficientsbeing differential expressions involving a partial <strong>de</strong>rivative of R(with a positive integer coefficient) and a partial <strong>de</strong>rivative of Q (with a negativeinteger coefficient). We have also:(7.28) ⎧R 3 = R x 3 + [3R x 2 u − Q x 3] U 1 + [3R xu 2 − 3Q x 2 u] (U 1 ) 2 +⎪⎨R 4 = R x 4 + [4R x 3 u − Q x 4] U 1 + [6R x 2 u 2 − 4Q x 3 u] (U 1 ) 2 ++ [4R xu 3 − 6Q x 2 u 2] (U1 ) 3 + [R u 4 − 4Q xu 3] (U 1 ) 4 + [−Q u 4] (U 1 ) 5 +⎪⎩161+ [R u 3 − 3Q xu 2] (U 1 ) 3 + [−Q u 3] (U 1 ) 4 + [3R xu − 3Q x 2] U 2 ++ [3R u 2 − 9Q xu ] U 1 U 2 + [−6Q u 2] (U 1 ) 2 U 2 + [−3Q u ] (U 2 ) 2 ++ [R u − 3Q x ] U 3 + [−4Q u ] U 1 U 3 .+ [6R x 2 u − 4Q x 3] U 2 + [12R xu 2 − 18Q x 2 u] U 1 U 2 ++ [6R u 3 − 24Q xu 2] (U 1 ) 2 U 2 + [−10Q u 3] (U 1 ) 3 U 2 ++ [3R u 2 − 12Q xu ] (U 2 ) 2 + [−15Q u 2] U 1 (U 2 ) 2 + [4R xu − 6Q x 2] U 3 ++ [4R u 2 − 16Q xu ] U 1 U 3 + [−10Q u 2] (U 1 ) 2 U 3 + [−10Q u ] U 2 U 3 ++ [R u − 4Q x ] U 4 + [−5Q u ] U 1 U 4 .Remark that all the brackets involved in equations (7.28) are of the form[λ R x a u − µ Q b+1 x a+1 ub], where λ, µ ∈ N and a, b ∈ N.In what follows we will not need the complete form of R κ but only thefollowing partial form:Lemma 8.1. For κ ≥ 4:(7.28)⎧R κ = R x κ + [ Cκ 1 R x κ−1 u − Q x κ]U 1 + [ Cκ 2 R x κ−2 u − Cκ 1 Q ]x κ−1 U 2 +⎪⎨⎪⎩+ [ C 2 κ R x 2 u − C 3 κ Q x 3 ]U κ−2 + [ C 1 κ R xu − C 2 κ Q x 2 ]U κ−1 ++ [ C 1 κ R u 2 − κ2 Q xu]U 1 U κ−1 + [ −C 2 κ+1 Q u]U 2 U κ−1 ++ [ R u − C 1 κ Q x]U κ + [ −C 1 κ+1 Q u]U 1 U κ ++ Remain<strong>de</strong>r,where the term Remain<strong>de</strong>r <strong>de</strong>notes the remaining terms in the expansion ofR κ .We note that the formula (7.28) is valid for κ = 3, comparing with (7.28),with the convention that the terms U κ−2 and U κ−1 vanish (they coinci<strong>de</strong> withU 1 and U 2 ), and replacing the coefficient −C 2 κ+1 Q u = −C 2 4 Q u = −6 Q uof the monomial U 2 U κ−1 by −3 Q u , as it appears in (7.28). The proof goesby a straightforward computation, applying the recursive <strong>de</strong>finition of thispartial formula.

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