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

54index l 2 . We get:(3.105)0 = −2G j y + l 1 2δj l 1G l 2y + l 2 Hj l 1 ,x − δj l 1H l 2l2 ,x ++ 2 ∑ G k L j l 1 ,k − ∑2δj l 1G k L l 2l2 ,k − ∑δj l 1G k L k k,k +kkka∑+ δ j l 1G k L k k,k − 1 ∑Hl k 2 1H j k + 1 ∑2 δj l 1Hl k 2H l 2k−kak∑ ∑− δ j l 1G k Θ k + δ j l 1G k Θ k + 1 2 δj l 1Θ 0 Θ 0 − 1kbkc 2 δj l 1Θ 0 Θ 0 .cbWe simplify, which yields the family (I) of partial differential relations ofTheorem 1.7 (3):k(3.106)0 = − 2 G j + 2 y l 1 δj l 1G l 2+ y l 2 Hj l 1 ,x − δj l 1H l 2l2 ,x +G k L j l 1 ,k − 2 ∑δj l 1G k L l 2l2 ,k ++ 2 ∑ kk− 1 ∑Hl k 2 1H j k + 1 ∑2 δj l 1kkH k l 2H l 2k.Secondly, let us insert inside (3.86) the values of Θ l 1 x , Θ l 2 x given by (3.101)and the value of Θ 0 given by (3.100). We place all the terms in the righthandside of the equality and we place the first order terms in the beginningy l 1(first three lines just below). We obtain:(3.107)0 = − 1 2 Hj l 1 ,y l 2 1+ L j l 1 ,l 2 ,x 4− 1 2 δj l 1L l 2l2 ,l 2 ,x5− 1 2 δj l 1L l 1l1 ,l 1 ,x6++ 1 3 δj l 1H l 2l 2 ,y l 2 2− 1 6 δj l 1L l 2l2 ,l 2 ,x5+ 1 3 δj l 2H l 1l 1 ,y l 1 3− 1 6 δj l 2L l 1l1 ,l 1 ,x6++ 1 3 δj l 1L l 2l2 ,l 2 ,x5− 1 6 δj l 1H l 2l 2 ,y l 2 2++ G j M l1 ,l 2 7 − 1 2− 1 4 δj l 2∑k− 2 3 δj l 1G l 2M l2 ,l 28∑kH j k Lk l 1 ,l 212+ 1 2∑Hl k 1L j l 2 ,kHl k 1Θ k + 1 4 δj l 2L l 1l1 ,l 1Θ 0 + 1c 4 δj l 2Θ 0 Θ l 1−db− 1 ∑3 δj l 1G k M l2 ,kkk1013+ 1 3 δj l 1∑k− 1 ∑4 δj l 2Hl k 1L k k,kkH l 2kL k l 2 ,l 214a−− 1 3 δj l 1∑kH k l 2L l 2l2 ,k15+