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

The induction formulas are⎧(3.4)⎪⎨Y i1 := D 1 i 1(Y ) −Y i1 ,i 2:= D 2 i 2(Y i1 ) −n∑ (D ) i 1 1 Xky k ,k=1n∑ (D ) i 1 2 Xky i1 ,k,k=1· · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · ·( )n∑ (⎪⎩ Y i1 ,i 2 ,...,i κ:= Di κ Yi1 κ ,i 2 ,...,i κ−1 − D ) i 1 κ Xky i1 ,i 2 ,...,i κ−1 ,k,are defined as in (1.22), drop-where the total differentiation operators Di λ ′ping the sums ∑ mj ′ =1 and the indices j′ .k=13213.5. Two instructing explicit computations. To begin with, let us computeY i1 . With Di 1 1= ∂ + y ∂∂x i 1 i 1, we have:∂yn∑ (Y i1 = D i1 (Y ) − Di 1 1 Xk 1)yk1(3.6)k 1 =1= Y x i 1 + Y y y i1 −n∑k 1 =1X k 1x i 1 y k 1−n∑k 1 =1X k 1y y i1 y k1 .Searching for formal harmony and for coherence with the formula (2.6) 1 ,we must include the term Y y y i1 inside the sum ∑ nk 1 =1 [·]y k 1. Using theKronecker symbol, we may write:n∑ [(3.7) Y y y i1 ≡ δk 1]i 1Y y yk1 .k 1 =1k 1 =1Also, we may rewrite the last term of (3.6) with a double sum:n∑n∑ [(3.8) − X k 1y y i1 y k1 ≡ −δk 1]i 1X k 2y yk1 y k2 .k 1 ,k 2 =1From now on and up to equation (3.39), we shall abbreviate any sum ∑ nk=1from 1 to n as ∑ k. Putting everything together, we get the final desiredperfect expression of Y i1 :(3.9) Y i1 = Y x i 1 + ∑ [δk 1]i 1Y y − X k 1x i yk1 + ∑ [1−δk 1]i 1X k 2y yk1 y k2 .k 1 k 1 ,k 2This completes the first explicit computation.The second one is about Y i1 ,i 2. It becomes more delicate, because severalalgebraic transformations must be achieved until the final satisfying formulais obtained. Our goal is to present each step very carefully, explaining every