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

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Next, we gather the un<strong>de</strong>rlined terms, or<strong>de</strong>ring them according to their number.This yields 6 collections of sums of monomials in the pure jet variab<strong>les</strong>:(3.18)Y i1 ,i 2= Y x i 1x i 2 + ∑ ][δ k 1i 1Y x i 2y + δ k 1i 2Y x i 1y − X k 1yx i 1x i 2 k1 +k 1+ ∑ ][δ k 1,k 2i 1 , i 2Y yy − δ k 1i 1X k 2x i 2y − δk 2i 2X k 1yx i 1y k1 y k2 +k 1 ,k 2+ ∑ [ ]−δ k 1,k 3i 1 , i 2X k 2yy y k1 y k2 y k3 +k 1 ,k 2 ,k 3+ ∑ k 1 ,k 2[δ k 1,k 2i 1 , i 2Y y − δ k 2i 2X k 1x i 1 − δk 2i 1X k 1x i 2]y k1 ,k 2++ ∑ []−δ k 2,k 3i 1 , i 2X k 1y − δ k 1,k 3i 1 , i 2X k 2y − δ k 1,k 3i 2 , i 1X k 2y y k1 y k2 ,k 3.k 1 ,k 2 ,k 3To attain the real perfect harmony, this last expression has still to be workedout a little bit.Lemma 3.19. The final expression of Y i1 ,i 2is as follows:(3.20)⎧Y i1 ,i 2= Y x i 1x i 2 + ∑ [δk 1i 1Y x i 2y + δ k 1i 2Y x i 1y − X k 1x i 1x 2] i yk1 +k 1⎪⎨⎪⎩+ ∑ ][δ k 1,k 2i 1 , i 2Y yy − δ k 1i 1X k 2x i 2y − δk 1i 2X k 2yx i 1y k1 y k2 +k 1 ,k 2+ ∑ [ ]−δ k 1,k 2i 1 , i 2X k 3yy y k1 y k2 y k3 +k 1 ,k 2 ,k 3+ ∑ k 1 ,k 2[δ k 1,k 2i 1 , i 2Y y − δ k 1i 1X k 2x i 2 − δk 1i 2X k 2x i 1]y k1 ,k 2++ ∑ []−δ k 1,k 2i 1 , i 2X k 3y − δ k 3,k 1i 1 , i 2X k 2y − δ k 2,k 3i 1 , i 2X k 1y y k1 y k2 ,k 3.k 1 ,k 2 ,k 3Proof. As promised, we explain every tiny <strong>de</strong>tail.The first lines of (3.18) and of (3.20) are exactly the same. For the transformationsof terms in the second, in the third and in the fourth lines, weuse the following <strong>de</strong>vice. Let Υ k1 ,k 2be an in<strong>de</strong>xed quantity which is symmetric:Υ k1 ,k 2= Υ k2 ,k 1. Let A k1 ,k 2be an arbitrary in<strong>de</strong>xed quantity. Thenobviously:∑(3.21)A k1 ,k 2Υ k1 ,k 2= ∑ A k2 ,k 1Υ k1 ,k 2.k 1 ,k 2 k 1 ,k 2Similar relations hold with a quantity Υ i1 ,i 2 ,...,i λwhich is symmetric withrespect to its λ indices. Consequently, in the second, in the third and in325

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