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

49the order of §3.73. We get:(3.89)12 δj l 1Θ l 2− 1 y l 32 δj l 1Θ l 3+ 1 y l 22 δj l 2Θ l 1− 1 y l 32 δj l 3Θ l 1= y l 2= L j l 1 ,l 2 ,y l 3 − Lj l 1 ,l 3 ,y l 2 + 1 2 δj l 1L l 3l 3 ,l 3 ,y l 2 − 1 2 δj l 1L l 2l 2 ,l 2 ,y l 3 ++ 1 2 δj l 3L l 1l 1 ,l 1 ,y l 2 − 1 2 δj l 2L l 1l 1 ,l 1 ,y l 3 ++ 1 2 Hj l 3M l1 ,l 2− 1 2 Hj l 2M l1 ,l 3+ 1 4 δj l 2L l 3l3 ,l 3L l 1l1 ,l 1− 1 4 δj l 3L l 2l2 ,l 2L l 1l1 ,l 1++ ∑ kL k l 1 ,l 3L j l 2 ,k − ∑ kL k l 1 ,l 2L j l 3 ,k ++ 1 ∑2 δj l 3L k l 1 ,l 2L k k,k − 1 ∑2 δj l 2L k l 1 ,l 3L k k,k +kk+ 1 4 δj l 2L l 1l1 ,l 1Θ l 3− 1 4 δj l 3L l 1l1 ,l 1Θ l 2+ 1 4 δj l 2L l 3l3 ,l 3Θ l 1− 1 4 δj l 3L l 2l2 ,l 2Θ l 1++ 1 ∑2 δj l 3L k l 1 ,l 2Θ k − 1 ∑2 δj l 2L k l 1 ,l 3Θ k +kk+ 1 2 δj l 2M l1 ,l 3Θ 0 − 1 2 δj l 3M l1 ,l 2Θ 0 + 1 4 δj l 2Θ l 3Θ l 1− 1 4 δj l 3Θ l 2Θ l 1.Next, replacing plainly (3.64) in (3.65) 4 , we get:(3.90) (Π00,0)y l 1 − ( Π 0 0,l 1)x == Θ 0 y l 1 − 1 2 Ll 1l1 ,l 1 ,x − 1 2 Θl 1 x == −Π 0 0,0 · Π0 l 1 ,0 − ∑ Π k 0,0 · a Π0 l 1 ,k + Π0 0,l 1 · Π 0 0,0 + ∑ akk= − ∑ (−G k) · (M l1 ,k) + ∑ (− 1 2 Hk l 1+ 1 )2 δk l 1Θ 0 ·kk( 1·2 Lk k,k + 1 )2 Θk == ∑ kG k M l1 ,k − 1 ∑Hl k 4 1L k k,k − 1 ∑4kkΠ k 0,l 1 · Π 0 0,k =H k l 1Θ k + 1 4 Ll 1l1 ,l 1Θ 0 + 1 4 Θ0 Θ l 1.